随笔分类 - leetcode
摘要：Counting Bits Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's in the binary re
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摘要：Word Pattern Given a pattern and a string s, find if s follows the same pattern. Here follow means a full match, such that there is a bijection betwee
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摘要：Add Digits Given an integer num, repeatedly add all its digits until the result has only one digit, and return it. Example 1: Input: num = 38 Output:
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摘要：somorphic Strings Given two strings s and t, determine if they are isomorphic. Two strings s and t are isomorphic if the characters in s can be replac
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摘要：Reverse String II Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string. If
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摘要：Base 7 Given an integer num, return a string of its base 7 representation. Example 1: Input: num = 100 Output: "202" Example 2: Input: num = -7 Output
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摘要：Fibonacci Number The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the
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摘要：Ransom Note Given two strings ransomNote and magazine, return true if ransomNote can be constructed by using the letters from magazine and false other
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摘要：Sum of Left Leaves Given the root of a binary tree, return the sum of all left leaves. A leaf is a node with no children. A left leaf is a leaf that i
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摘要：Assign Cookies Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each
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摘要：Fizz Buzz Given an integer n, return a string array answer (1-indexed) where: answer[i] == "FizzBuzz" if i is divisible by 3 and 5. answer[i] == "Fizz
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摘要：Valid Anagram Given two strings s and t, return true if t is an anagram of s, and false otherwise. An Anagram is a word or phrase formed by rearrangin
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摘要：Remove Linked List Elements Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and
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摘要：Move Zeroes 思路一: 用 left 指针标记求解数组最大下标 + 1，初始化的时候是 0，随着往右遍历，left 会一步步扩大。最后把 left 右边的数都置为 0。 这题的关键在于 left 永远指向求解数组最大下标 + 1 public void moveZeroes(int[] n
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摘要：Power Of Two 思路一: 观察 2 的 n 次方的二进制，都只有一位 1 bit，遍历即可 public boolean isPowerOfTwo(int n) { if (n <= 0) return false; int count = 0; for (int i = 0; i < 3
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摘要：Majority Element 思路一: map public int majorityElement(int[] nums) { if (nums.length == 1) return nums[0]; Map<Integer, Integer> map = new HashMap<>();
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摘要：Reverse Bits 思路一: 遍历 32 位 bit，记录 bit 结果 public int reverseBits(int n) { int result = 0; int x = 32; while (x-- > 0) { int bit = n & 1; result <<= 1; i
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摘要：Long Pressed Name 思路一: 暴力，遍历两个字符串，对比。边界情况不太好处理 public boolean isLongPressedName(String name, String typed) { if (typed.length() < name.length()) retur
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摘要：X of a Kind in a Deck of Cards 思路一: 统计数字个数，然后用 [2, length] 的约数尝试所有数字个数，判断能否整除 public boolean hasGroupsSizeX(int[] deck) { Map<Integer, Integer> map =
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摘要：Monotonic Array 思路一: 遍历 public boolean isMonotonic(int[] nums) { boolean increase = true; for (int i = 0; i < nums.length - 1; i++) { if (nums[i] > nu
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