思路: dp[i][j]表示到第i + 1个位置为止,并且以剩下的所有数字中第j + 1小的数字为结尾所有的合法序列数. 实现: class Solution { public: int numPermsDISequence(string S) { ; ; vector<vector<, vector<, )); ; i <= n; i++) dp[][i] = ; ; i <= n; i++) { ] == 'D') { dp[i & ][n - i] = dp[i…
We are given S, a length n string of characters from the set {'D', 'I'}. (These letters stand for "decreasing" and "increasing".) A valid permutation is a permutation P[0], P[1], ..., P[n] of integers {0, 1, ..., n}, such that for all …
We are given S, a length n string of characters from the set {'D', 'I'}. (These letters stand for "decreasing" and "increasing".) A valid permutation is a permutation P[0], P[1], ..., P[n] of integers {0, 1, ..., n}, such that for all…
We are given S, a length n string of characters from the set {'D', 'I'}. (These letters stand for "decreasing" and "increasing".) A valid permutation is a permutation P[0], P[1], ..., P[n] of integers {0, 1, ..., n}, such that for all …
We are given S, a length n string of characters from the set {'D', 'I'}. (These letters stand for "decreasing" and "increasing".) A valid permutation is a permutation P[0], P[1], ..., P[n] of integers {0, 1, ..., n}, such that for all …
Given N pairs of parentheses “()”, return a list with all the valid permutations. Assumptions N > 0 Examples N = 1, all valid permutations are ["()"] N = 3, all valid permutations are ["((()))", "(()())", "(())()"…
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All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017) .   Top Interview Questions # Title Difficulty Acceptance 1 Two Sum Medium 17.70% 2 Add Two N…
# Title Solution Acceptance Difficulty Frequency     4 Median of Two Sorted Arrays       27.2% Hard     10 Regular Expression Matching       25.6% Hard     23 Merge k Sorted Lists       35.8% Hard     25 Reverse Nodes in k-Group       37.7% Hard    …
链接:https://leetcode.com/tag/divide-and-conquer/ [4]Median of Two Sorted Arrays [23]Merge k Sorted Lists [53]Maximum Subarray (2019年1月23日, 谷歌tag复习) 最大子段和. 题解: follow up 是divide and conquer If you have figured out the O(n) solution, try coding another…