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Leetcode第1题至第10题 思路分析及C++实现

笔者按照目录刷题,对于每一道题,力争使用效率最高(时间复杂度最低)的算法,并全部通过C++代码实现AC。(文中计算的复杂度都是最坏情况复杂度)
因为考虑到大部分读者已经在Leetcode浏览过题目了,所以每道题都按照 解题思路 -> 实现代码 -> 问题描述 的顺序进行讲解。
(笔者目前已刷 40 题,已更新解法 10 题,最近一段时间会频繁更新)可以点击下方链接,直达gitbook:
https://codernie.gitbooks.io/leetcode-solutions/content/
也欢迎大家关注我的微信公众号(大雄的学习人生),有问题可以随时后台回复我,多多探讨。

目录:

1 Two Sum

2 Add Two Numbers

3 Longest Substring Without Repeating Characters

4 Median of Two Sorted Arrays

5 Longest Palindromic Substring

6 ZigZag Conversion

7 Reverse Integer

8 String to Integer (atoi)

9 Palindrome Number

10 Regular Expression Match

1. Two Sum 【easy】

解题思路

暴力解法 O(N^2):

嵌套两层循环:第一层:i 从 0 到 n - 2;第二层:j 从 i + 1 到 n - 1;判断 nums[i] + nums[j] == target ,如果成立则是正确答案

map解法 O(N*logN)

从 0 到 n - 1 依次遍历,利用map存放每一个数值的下标,在map中寻找是否有使(nums[i] + x == target)成立的x的存在,如果存在则返回i和它的下标(即myMap[ target - nums[i] ])。

复杂度分析:因为只遍历了一次数组,map每次的查询的时间复杂度为O(logN)所以整体复杂度为O(N*logN)、如果这里使用hash_map可以将查询复杂度降低到O(1),从而使得整体复杂度为O(N),但是hash_map不是标准的C++库,所以这里没有使用。

实现代码:

// 1. Two Sum
vector<int> twoSum(vector<int>& nums, int target) {
  map<int, int> myMap;
  vector<int> result;
  for (int i = 0; i < nums.size(); i++) {
    if (myMap.find(target - nums[i]) == myMap.end()) {
      myMap[nums[i]] = i;
    } else {
      result = {myMap[target - nums[i]], i};
      break;
    }
  }
  return result;
}

问题描述:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have _exactly _one solution, and you may not use the_same_element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

2. Add Two Numbers【medium】

解题思路

从左到右遍历链表,依次相加,每一个位置生成一个新的结点即可。

时间复杂度:O( max( len(l1), len(l2) ) )

考虑边界条件:

1.进位的的处理:carry表示进位,当最后一位还有进位时,即使 l1 和 l2 均为NULL的情况下,还需要生成一个新的结点,所以while的条件中加入了 carry != 0 判断项。

2.返回头结点:当头结点为NULL的时候记录头结点,并且让p等于头结点;后续情况让 p->next 等于新的结点,并让 p 指向 p->next。

实现代码:

// 2. Add Two Numbers
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
  ListNode *head = NULL, *p;
  int carry = 0, sum;
  while (l1 != NULL || l2 != NULL || carry != 0) {
    sum = 0;
    if (l1 != NULL) {
      sum += l1->val;
      l1 = l1->next;
    }
    if (l2 != NULL) {
      sum += l2->val;
      l2 = l2->next;
    }
    sum += carry;
    carry = sum / 10;
    sum %= 10;
    ListNode *newNode = new ListNode(sum);
    if (head == NULL) {
      head = newNode;
      p = newNode;
    } else {
      p->next = newNode;   
      p = p->next;
    }
  }
  return head;
}

问题描述:

You are given two non-empty linked lists representing two non-negative integers. The digits are stored inreverse orderand each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

3. Longest Substring Without Repeating Characters【medium】

解题思路

暴力解法 O(N^3):

从每个点遍历、依次遍历这个点后的每一个点、通过遍历该点之前的点判断该点是否出现过。听上去有点拗口,代码在下方,这个暴力方法Leetcode也可以AC,但是不推荐使用。

头尾标记法 O(N*logN):

头标记指向当前最长无重复字符串的头部,尾标记指向其尾部。通过一个 map 来记录出现过的字符最后出现的位置。

依次遍历数组,如果当前字符已出现过,则让头标记指向其最后出现过的位置的后一个位置。然后每次通过头、尾标记计算当前无重复字符串的长度,并与已知最大值作比较。这里查询map的复杂度为 O(logN),遍历的复杂度为 O(N),因此整体复杂度为 O(N*logN)。如果这里使用hash_map可以将查询复杂度降低到O(1),从而使得整体复杂度为O(N),但是hash_map不是标准的C++库,所以这里没有使用。

实现代码:

// 3. Longest Substring Without Repeating Characters
// 暴力解法
int lengthOfLongestSubstring_bruteForce(string s) {
  int res = 0, sum;
  for (int i = s.size() - 1; i >= 0; i--) {
    sum = 1;
    for (int j = i - 1; j >= 0; j--) {
      bool flag = true;
      for (int k = i; k > j; k--) {
        if (s[j] == s[k]) {
          flag = false;
          break;
        }
      }
      if (flag) {
        sum++;
      } else {
        break;
      }
    }
    res = max(res, sum);
  }
  return res;      
}
// 头尾标记法
int lengthOfLongestSubstring(string s) {
  map<char, int> myMap;
  int res = 0;
  for (int i = 0, j = 0; j < s.size(); j++){
    if (myMap.find(s[j]) != myMap.end()) {
      i = max(i, myMap[s[j]] + 1);
    }
    myMap[s[j]] = j;
    res = max(res, j - i + 1);
  }
  return res;
}

问题描述:

Given a string, find the length of thelongest substringwithout repeating characters.

Examples:

Given"abcabcbb", the answer is"abc", which the length is 3.

Given"bbbbb", the answer is"b", with the length of 1.

Given"pwwkew", the answer is"wke", with the length of 3. Note that the answer must be asubstring,"pwke"is asubsequenceand not a substring.

4. Median of Two Sorted Arrays【hard】

解题思路

这道题咋一看像二分查找,但是仔细看题,发现有两个有序数组,而且不是让我们找一个特定的数,而是要找两个数组合并后的中位数,这样一看就比较难了,也难怪归类为hard类别。这道题除了一下介绍的二分查找法,还有两个数组分别进行二分查找的方法,不过代码量相对更多(也可能是因为笔者水平不够导致代码量过大),而且看了下面的二分查找法后,感叹于该算法作者的脑洞,所以在这里只介绍该种方法。

暴力方法 O((m + n)*log(m + n)):

将两个数组合并,然后进行快排,中间的数即中位数。由于题目说了复杂度不能超过O(log(m + n)),所以这个方法当然回Time Limit Excess,所以我们得探究一种更高效的解法。

二分查找法 O(log(min(m, n))):

首先分别把两个数组分成两边,大概为下面这种形式:(A表示nums1, B表示nums2)

          left_part          |        right_part
    A[0], A[1], ..., A[i-1]  |  A[i], A[i+1], ..., A[m-1]
    B[0], B[1], ..., B[j-1]  |  B[j], B[j+1], ..., B[n-1]

因为有序数列的性质,我们知道只要我们满足以下两个条件,我们就可以找到中位数了:

条件一:len(A_left) + len(B_left) = len(A_right) + len(B_right)

条件二:max[A_left, B_left] <= min[A_right, B_right]

为了使问题简单化,我们先只考虑 m + n 为偶数的情况下,只要满足上述两个条件,我们的中位数就等于左边的最大值加上右边的最小值除以二。

为了满足条件一,我们只要令** i + j == m + n - i - j (+ 1)**即可(这里加一是为了之后考虑 m + n 为奇数的情况)

而为了满足条件二,根据有序数组的性质,我们知道只需要满足 A[i - 1] <= B[j] 且 B[j - 1] <= A[i] 即可。

接下来开始我们的算法探究:

假设我们首先随机选择一个 i (这里 0 <= i < m),所以我们根据条件一,可以求得 j = (m + n + 1) / 2 - i;

为了满足条件二,我们开始分别比较 A[i - 1] 与 B[j] 和 B[j - 1] 与 A[i]:

不难知道可能会有四种情况:

  • A[i - 1] <= B[j] 且 B[j - 1] <= A[i] :这不正是我们要找的 i 吗?可以直接返回答案
  • A[i - 1] > B[j] 且 B[j - 1] > A[i] :根据有序数列的性质,再利用反证法不难证明这种情况不可能存在
  • A[i - 1] <= B[j] 且 B[j - 1] > A[i]:为了使情况更加接近我们的答案,也就是情况1。也就是要使 B[j - 1] <= A[i],因为现在 B[j - 1] > A[i],所以我们要想办法缩小 B[j - 1],扩大A[i],所以当然是让我们的 i 增大,否则情况会越来越远离我们的正确答案。
  • A[i - 1] > B[j] 且 B[j - 1] <= A[i]:与情况3恰恰相反,这里我们应该缩小我们的 i。

那我们如何缩小和扩大我们的 i 呢,那就是采用二分查找的方式啦,首先将 i 置为数组A的中间下标,如果需要增大,则把其设为上半区的中间下标,反之则设为下半区的中间下标,所以这种搜索方式的时间复杂度和二分查找的时间复杂度一样,为了使时间复杂度尽量的小,我们使A成为长度更小的那个数组,如果初始A比B长,我们则交换它们的位置。

考虑边界条件:

1.如果不存在满足条件二的情况会怎么样呢?也就是 i 走到了数组A的尽头,依旧没法满足条件二,这个时候我们不难知道如果 i 走到数组A的最左端,那么它一定是在不断地经历情况4,而这时 A[0] > B[j],那么我们不难知道这时left_part的最大值就是B[j - 1];反之我们也可以推出 i 到了A的最右端、j 到了B的最左端或者最右端的情况。

2.m + n 为奇数。这个时候我们不难推出 left_part 的最大值就是中位数。

3.A或B为空数组,因为当数组空的时候无法对数组进行下标访问,所以我们在进行二分查找前就应该对该情况进行特殊处理,处理方式也是很简单的啦。

实现代码:

// 4. Median of Two Sorted Arrays
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
  // 使得 nums1 短于 nums2
  int m = nums1.size();
  int n = nums2.size();
  if (m > n) {
    vector<int> temp = nums1;
    nums1 = nums2;
    nums2 = temp;
    m = m + n;
    n = m - n;
    m = m - n;
  }
  // 考虑数组长度为0的边界情况
  if (m == 0) {
    if (n == 0) {
      return 0;
    } else {
      if (n % 2 == 1) {
        return nums2[n / 2];
      } else {
        return (double)(nums2[n / 2] + nums2[n / 2 - 1]) / 2;
      }
    }
  }
  int iMin = 0, iMax = m, sizeSum = (m + n + 1) / 2, i, j;
  while (iMin <= iMax) {
    i = (iMax + iMin) / 2;
    j = sizeSum - i;
    if (nums2[j - 1] > nums1[i] && i < iMax) {
      iMin = i + 1;
    } else if (nums1[i - 1] > nums2[j] && i > iMin) {
      iMax = i - 1;
    } else {
      int maxLeft, minRight;
      if (i == 0) {
        maxLeft = nums2[j - 1];
      } else if (j == 0) {
        maxLeft = nums1[i - 1];
      } else {
        maxLeft = max(nums1[i - 1], nums2[j - 1]);
      }
      if ((m + n) % 2 == 1) {
        return maxLeft;
      }
      if (i == m) {
        minRight = nums2[j];
      } else if (j == n) {
        minRight = nums1[i];
      } else {
        minRight = min(nums1[i], nums2[j]);
      }
      return (double)(maxLeft + minRight) / 2;
    }
  }
  return 0;
}

问题描述:

There are two sorted arrays nums1 and nums2 of size m and n respectively.

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

5. Longest Palindromic Substring【medium】

解题思路

暴力解法 O(N^3):

字符串有 n(n-1)/2 个子串,对每个子串进行检测,看其是否是回文子串。因此复杂度为 O(N^3)。

从字符串中间开始扩展 O(N^2):

把字符串的每个点分别当成回文子串的中间点,开始往两端扩展,不断检测,直到两端不相等为止。因此复杂度为 O(N^2)。

动态规划 O(N^2):

用 dp[i][j] 表示下标为 i 开头 j 结尾的子串是否是回文子串。

转移方程:dp[i][j] = (dp[i + 1][j - 1] && s[i] == s[j]) 【含义:当且仅当子串首尾两端相等,且去除首尾两端依旧是回文串时,该子串才会是回文串】

初始条件:对于每个长度为1的子串 dp[i][i] 都为回文串;对于每个长度为2的子串 dp[i][i + 1],当其首尾两端相等时,其为回文串,否则不是。

实现代码:

// 5. Longest Palindromic Substring (动态规划)
string longestPalindrome(string s) {
  int length = s.size();
  if (length == 0) return s;
  int resI = 0, resJ = 0;
  bool dp[length + 1][length + 1];
  for (int i = 0; i <= length; i++)
    dp[i][i] = true;
  for (int i = 0; i < length; i++) {
    if (s[i] == s[i + 1]) {
      dp[i][i + 1] = true;
      if (resJ - resI < 1) {
        resI = i;
        resJ = i + 1;
      }
    } else {
      dp[i][i + 1] = false;
    }
  }
  for (int gap = 2; gap < length; gap++) {
    for (int i = 0; i + gap < length; i++) {
      int j = i + gap;
      if (s[i] == s[j] && dp[i + 1][j - 1]) {
        dp[i][j] = true;
        if (resJ - resI < j - i) {
          resI = i;
          resJ = j;
        }
      } else {
        dp[i][j] = false;
      }
    }
  }
  return s.substr(resI, resJ - resI + 1);
}

问题描述:

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: "cbbd"
Output: "bb"

6. ZigZag Conversion【medium】

解题思路

这道题倒没有特别的方法,就按照题目意思来模拟 Z 字形即可,用一个字符串数组来存放每一行的字符串,最后进行拼接即可。

考虑边界条件:

当numRows等于1的时候,因为point无法增加也无法减小,所以没办法共用后面的代码,考虑到numRows等于1的时候,答案就是原字符串,所以这里直接返回s即可。

实现代码:

// 6. ZigZag Conversion
string convert(string s, int numRows) {
  if (numRows == 1) return s;
  string res;
  bool shouldIncrease = true;
  string strArr[numRows];
  int point = 0;
  for (char c : s) {
    strArr[point] += c;
    if (point == numRows - 1) {
      shouldIncrease = false;
    } else if (point == 0) {
      shouldIncrease = true;
    }
    if (shouldIncrease) {
      point++;
    } else {
      point--;
    }
  }
  for (string str: strArr) {
    res += str;
  }
  return res;
}

问题描述:

The string"PAYPALISHIRING"is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line:"PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"

Explanation:
P     I    N
A   L S  I G
Y A   H R
P     I

7. Reverse Integer【easy】

解题思路

挨个遍历,不断把末位数赋给新的值即可。

考虑边界条件:

当结果溢出时返回0,所以为了不让中间值溢出,采用 long 类型来保存结果。

实现代码:

// 7. Reverse Integer
int reverse(int x) {
  long result = 0, longX = abs((long)x);
  while (longX > 0) {
    result = result * 10 + longX % 10;
    longX /= 10;
  }
  result = (x > 0) ? result : -result;
  if (result > INT32_MAX || result < INT32_MIN) {
    return 0;
  } else {
    return (int)result;
  }
}

问题描述:

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

8. String to Integer (atoi)【medium】

解题思路

遍历字符串然后进行分情况讨论:( isInit 表示数字是否已经开始,通过 isInit 的值判断是否为开头,如果为 true 表示不是开头)

(1) 空格:如果为开头空格则continue,否则跳出循环

(2) 正负号:如果为开头正负号则设置isNeg的值,否则跳出循环

(3) 数字:将 isInit 置为true,累加结果

(4) 其他符号:跳出循环

考虑边界条件:

当结果溢出时根据正负返回 INT32_MAX 或者 INT32_MIN,所以为了不让中间值溢出,采用 long 类型来保存结果。

实现代码:

// 8. String to Integer (atoi)
int myAtoi(string str) {
  long result = 0;
  bool isInit = false;
  bool isNeg = false;
  for (char c : str) {
    if (c == ' ') {
      if (isInit) {
        break;
      } else {
        continue;
      }
    } else if (c == '-' || c == '+') {
      if (!isInit) {
        isInit = true;
      } else {
        break;
      }
      isNeg = (c == '-');
    } else if (c >= 48 && c <= 57) {
      isInit = true;
      result = result * 10 + (c - 48);
      if (result > INT32_MAX) { 
        return isNeg ? INT32_MIN : INT32_MAX;
      }
    } else {
      break;
    }
  }      
  return (int)(isNeg ? -result : result);
}

问题描述:

Implementatoiwhich converts a string to an integer.

The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned.

Note:

  • Only the space character' 'is considered as whitespace character.
  • Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. If the numerical value is out of the range of representable values, INT_MAX (2^31 − 1) or INT_MIN (−2^31) is returned.

Example 1:

Input: "42"
Output: 42

Example 2:

Input: "   -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
             Then take as many numerical digits as possible, which gets 42.

Example 3:

Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.

Example 4:

Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical 
             digit or a +/- sign. Therefore no valid conversion could be performed.

Example 5:

Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
             Thefore INT_MIN (−231) is returned.

9. Palindrome Number【medium】

解题思路

利用第七题的代码,将数字反转,判断与原数字是否相等即可,这里考虑到负数全部都不是回文数字,所以直接返回false。

实现代码:

// 9. Palindrome Number
int reverse(int x) {
  long result = 0, longX = abs((long)x);
  while (longX > 0) {
    result = result * 10 + longX % 10;
    longX /= 10;
  }
  result = (x > 0) ? result : -result;
  if (result > INT32_MAX || result < INT32_MIN) {
    return 0;
  } else {
    return (int)result;
  }
}
bool isPalindrome(int x) {
  if (x < 0) {
    return false;
  } else {
    return (x == reverse(x));
  }
}

问题描述:

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121
Output: true

Example 2:

Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Follow up:

Coud you solve it without converting the integer to a string?

10. Regular Expression Matching【hard】

解题思路

动态规划思想解答这道题:

用 dp[i][j] 表示 s 的前 i 个字符组成的字符串和 p 的 前 j 个字符组成的字符串是否匹配。

转移方程:

当 p[j - 1] == '*' 时:因为 * 可以表示匹配零位或者多位,正则匹配这里要做贪心考虑,分三种情况,只要其中一种满足即为true:

  • 匹配零位:则 dp[i][j] = dp[i][j - 2]
  • 匹配一位:则 dp[i][j] = dp[i - 1][j - 2] && (满足最后一位匹配)
  • 匹配多位:则一位一位匹配,dp[i][j] = dp[i - 1][j] && (满足最后一位匹配)

当 p[j - 1] != '*' 时,dp[i][j] 当且仅当 dp[i - 1][j - 1]为true时,并且最后一位匹配成功时,才为true。

初始状态:

显然,当 s 不为空,p 为空的时候dp[i][j] = false;

其次,当 s 为空,p不为空的时候,考虑到 * 可以匹配零位,所以利用状态转移方程判断其是否应该为true。

实现代码:

// 10. Regular Expression Matching
bool isMatch(string s, string p) {
  int n = s.size();
  int m = p.size();
  // initial
  bool dp[n + 1][m + 1];
  for (int i = 0; i < n + 1; i++) {
    for (int j = 0; j < m + 1; j++) {
      dp[i][j] = false;
    }
  }
  // start
  dp[0][0] = true;
  for (int i = 1; i < n + 1; i++) {
    dp[i][0] = false;
  }
  for (int j = 1; j < m + 1; j++) {
    if (j % 2 == 0) {
      dp[0][j] = dp[0][j - 2] && p[j - 1] == '*';
    } else {
      dp[0][j] = false;
    }
  }
  // trans
  bool compare;
  for (int i = 1; i < n + 1; i++) {
    for (int j = 1; j < m + 1; j++) {
      if (p[j - 1] != '*') {
        dp[i][j] = dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
      } else {
        compare = (s[i - 1] == p[j - 2] || p[j - 2] == '.');
        dp[i][j] = dp[i][j - 2] || (dp[i - 1][j - 2] && compare) || (dp[i - 1][j] && compare);
      }
    }
  }
  return dp[n][m];
}

问题描述:

Given an input string (s) and a pattern (p), implement regular expression matching with support for'.'and'*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover theentireinput string (not partial).

Note:

  • scould be empty and contains only lowercase lettersa-z
  • pcould be empty and contains only lowercase lettersa-z, and characters like . or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false
posted @ 2018-06-13 21:59 大雄的学习人生 阅读(...) 评论(...) 编辑 收藏
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