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用split() 不用API, better solution, O(N) time O(1) space 阅读全文
posted @ 2016-12-07 00:36
neverlandly
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refer to https://discuss.leetcode.com/topic/63854/o-n-java-solution-by-analyzing-changes-allowed-to-fix-each-problem http://www.cnblogs.com/grandyang/ 阅读全文
posted @ 2016-12-06 05:38
neverlandly
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真是被这道题气死了，总是弄不对 The problem says that we can make at most k changes to the string (any character can be replaced with any other character). So, let's 阅读全文
posted @ 2016-12-05 13:58
neverlandly
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# of '0': # of 'z' # of '2': # of 'w' 4: u 6: x 8: g 3: h - 8 5: f - 4 7: s - 6 1: o - 0 - 2 - 4 9: i - 5 - 6 - 8 我的code用了一个数组来存char count 阅读全文
posted @ 2016-12-05 08:51
neverlandly
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Solution 1: Bit Manipulation: 这题真心有点难，get the max XOR bit by bit note that if A^B=C, then A^C=B, B^C=A Solution 2: Trie, (未研究) https://discuss.leetcod 阅读全文
posted @ 2016-12-05 07:48
neverlandly
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只计算每个battleship的第一个元素，所以后面‘X’如果above或者left也是'X'的话，不被计算 阅读全文
posted @ 2016-12-05 06:10
neverlandly
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只有注册用户登录后才能阅读该文。 阅读全文
posted @ 2016-12-05 04:44
neverlandly
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Time Complexity will be O(n) because the "start" and "end" points will only move from left to right once. Sliding Window: Use a count to denote the di 阅读全文
posted @ 2016-12-03 13:14
neverlandly
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这题考点在于需要设置两个visited数组 Two Queue and add all the Pacific border to one queue; Atlantic border to another queue. Keep a visited matrix for each queue. I 阅读全文
posted @ 2016-12-03 12:05
neverlandly
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Backpack problem: dp[i][j] means if the first i elements can sum up to value j dp[i][j] = dp[i-1][j] || (j>=nums[i-1] && dp[i-1][j-nums[i-1]]) the res 阅读全文
posted @ 2016-12-03 10:29
neverlandly
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My Solution: Highest votes in discussion 阅读全文
posted @ 2016-12-03 06:22
neverlandly
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A slice (P, Q)其实就是从P到Q的序列，（P，Q）是arithmetic slice要求是等差数列且至少有三个数，问array里面有多少个这样的（P, Q） 这道题难点在于想到用DP，一旦想到用DP，就迎刃而解 dp[i]表示以i结束的slice有多少个 阅读全文
posted @ 2016-12-03 05:12
neverlandly
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The above image represents the elevation map [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]] before the rain. After the rain, water are trapped between th 阅读全文
posted @ 2016-12-03 03:16
neverlandly
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Greedy + Stack: 用一个栈维护最后要留存下来的digits 需要注意的是：如果遍历到string的某个char, string后面的char数刚刚好能填满这个栈，那么即使这个char比栈顶还要小，也不出栈，直接入栈 最后要删除leading 0 用char[]实现栈，思路一样，要快很多 阅读全文
posted @ 2016-12-03 00:24
neverlandly
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Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2. Note: The length of both num1 and num2 is =0 || j>=0 || carry!=0) { 8 int sum = 0; ... 阅读全文
posted @ 2016-12-02 12:46
neverlandly
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Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters. This is case sensitive, for example "Aa" is not cons... 阅读全文
posted @ 2016-12-02 12:26
neverlandly
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refer to: https://discuss.leetcode.com/topic/60394/easy-concept-with-python-c-java-solution E.g.input: [[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]subar 阅读全文
posted @ 2016-12-02 11:58
neverlandly
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记住要delete begining 0 要记住单独处理 num == 0的情况 阅读全文
posted @ 2016-12-02 11:06
neverlandly
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Solution 1: Backtracking, but will TLE Better Solution: DP (refer to https://discuss.leetcode.com/topic/59903/very-easy-to-understand-java-solution-wi 阅读全文
posted @ 2016-12-02 08:35
neverlandly
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Solution 1: Bit Manipulation use Integer.bitCount() Solution 2: Backtracking, 非常精妙之处在于用了两个数组来帮助generate digit（例如：1011 -> 11） 阅读全文
posted @ 2016-12-02 05:00
neverlandly
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Graph, DFS (1) Build the map, the key is dividend, the value is also a map whose key is divisor and value is its parameter. For example, a / b = 2.0, 阅读全文
posted @ 2016-12-02 02:10
neverlandly
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Solution 1: Solution with out using % 阅读全文
posted @ 2016-12-01 12:46
neverlandly
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Refer to: https://discuss.leetcode.com/topic/58334/a-couple-of-java-solutions-with-explanations/2 When to add 1 instead of minus 1, here is an example 阅读全文
posted @ 2016-12-01 12:31
neverlandly
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Analysis: Given a string s, find out all chars that are invalid (i.e., count < k). The longest substring must reside in one of the substrings divided 阅读全文
posted @ 2016-12-01 11:33
neverlandly
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F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25 [4, 3, 2, 6] F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16 [6, 阅读全文
posted @ 2016-12-01 08:12
neverlandly
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这道题题干给出了判断 one single UTF-8 char的方法，然后给一个UTF-8 char sequence，判断是不是正确sequence. (读题读了很久) 这道题关键是要学到用 & 取出一个bit sequence当中几位的方法 二进制数表示法：在前面加 0b, 八进制加0o, 十 阅读全文
posted @ 2016-12-01 07:09
neverlandly
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自己的做法：这种括号问题肯定是用栈，最好是先在栈里存一个空元素，然后stack.peek().append()各种操作 一个栈存string, 一个栈存number， 维护一个指针numStart指向数字的开始 1. 遇到数字啥也不做 2. 遇到char： stack.peek().append(c 阅读全文
posted @ 2016-12-01 05:07
neverlandly
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Refer to https://discuss.leetcode.com/topic/56052/really-easy-understanding-solution-o-n-java and https://discuss.leetcode.com/topic/55923/o-n-solutio 阅读全文
posted @ 2016-12-01 02:12
neverlandly
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Three types of answer: Map Solution, O(N) memory, O(N) init, O(1) pick. Like @dettier's Reservoir Sampling. O(1) init, O(1) memory, but O(N) to pick. 阅读全文
posted @ 2016-11-30 12:23
neverlandly
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Recursion: 是不是left子数完全由bottom往上第二层决定，如果是left子树且是叶子节点，那么就是left leaves, parent得告诉child是不是在left子树 BFS: 阅读全文
posted @ 2016-11-30 11:45
neverlandly
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Basic Solution: DP, O(mn) time, O(m) space, m is the size of s, n is the size of t Greedy Solution: O(n) time, O(1) space Follow Up: The best solution 阅读全文
posted @ 2016-11-30 11:09
neverlandly
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refer to https://discuss.leetcode.com/topic/59293/java-easiest-solution-o-logn-with-explanation Time Complexity: O(log n) update and record head in ea 阅读全文
posted @ 2016-11-30 06:52
neverlandly
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O（N）time, O(1) space 阅读全文
posted @ 2016-11-30 05:40
neverlandly
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Time Complexity: O(N) The depth of the directory/file is calculated by counting how many "\t"s are there.The time complexity is O(n) because each subs 阅读全文
posted @ 2016-11-30 05:19
neverlandly
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有括号这种一般要用stack, stack top 就是当前着眼的那一个NestedInteger, 可以对其添加新的元素。 注意47行判断很关键，顺带处理了 "[]," 括号后面是逗号的情况 阅读全文
posted @ 2016-11-30 02:28
neverlandly
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Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples: s = "leetcode" return 0. s = "loveleetcode", return 2. Note: You may a... 阅读全文
posted @ 2016-11-29 12:24
neverlandly
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Solution 1: Solution 2: O(N) time, O(1) space The basic idea is to find the next number to add.Take 45 for example: if the current number is 45, the n 阅读全文
posted @ 2016-11-29 11:59
neverlandly
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Random random = new Random(); random.nextInt(int i); 阅读全文
posted @ 2016-11-29 10:53
neverlandly
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using one array 阅读全文
posted @ 2016-11-29 06:41
neverlandly
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Solution 1: Reservior sampling: (wiki introduction) Reservoir sampling is a family of randomized algorithms for randomly choosing a sample of k items 阅读全文
posted @ 2016-11-29 06:19
neverlandly
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