ArgenBarbie - 博客园

2016年3月5日

摘要： 1. 每个元素可用多次 Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The 阅读全文

posted @ 2016-03-05 18:17 ArgenBarbie 阅读(205) 评论(0) 推荐(0)

37. Sudoku Solver *HARD*

摘要： Write a program to solve a Sudoku puzzle by filling the empty cells. Empty cells are indicated by the character '.'. You may assume that there will be 阅读全文

posted @ 2016-03-05 16:32 ArgenBarbie 阅读(318) 评论(0) 推荐(0)

36. Valid Sudoku

摘要： Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku board could be partially filled, where empty cells are filled wit 阅读全文

posted @ 2016-03-05 16:17 ArgenBarbie 阅读(162) 评论(0) 推荐(0)

33. 81. Search in Rotated Sorted Array *HARD*

摘要： Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target va 阅读全文

posted @ 2016-03-05 14:43 ArgenBarbie 阅读(200) 评论(0) 推荐(0)

32. Longest Valid Parentheses *HARD*

摘要： Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring. For "(()", the lo 阅读全文

posted @ 2016-03-05 12:01 ArgenBarbie 阅读(179) 评论(0) 推荐(0)

31. Next Permutation

摘要： Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possib 阅读全文

posted @ 2016-03-05 11:12 ArgenBarbie 阅读(165) 评论(0) 推荐(0)

30. Substring with Concatenation of All Words *HARD*

摘要： You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a conca 阅读全文

posted @ 2016-03-05 10:24 ArgenBarbie 阅读(211) 评论(0) 推荐(0)

2016年3月4日

29. Divide Two Integers

摘要： Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. (1) log int divide(int dividend, int d 阅读全文

posted @ 2016-03-04 23:40 ArgenBarbie 阅读(216) 评论(0) 推荐(0)

2016年3月3日

KMP算法

摘要： void getNext(string needle, int *next) { int l = needle.length(), i, k; next[0] = -1; k = -1; for(i = 0; i < l-1;) { if(-1 == k || needle[i] == needle 阅读全文

posted @ 2016-03-03 23:36 ArgenBarbie 阅读(205) 评论(0) 推荐(0)

23. Merge k Sorted Lists

摘要：合并k个有序列表。（1）堆排序 bool cmp(ListNode *l1,ListNode *l2){ return l1->val > l2->val; } class Solution { public: ListNode* mergeKLists(vector<ListNode*>& li 阅读全文

posted @ 2016-03-03 17:26 ArgenBarbie 阅读(178) 评论(0) 推荐(0)