皓波的博客

总共用了两种解法，一种是我自己递归求解的，效率很低，边界条件很难理清楚，代码很长，第二种是DP解法，把递归的子问题的解存到一个数组里，代码很短。 1. 递归解法 主要想法是根据当前的正则表达式，来决定接下去如何匹配，我们可以用两个字符串的子串去匹配，麻烦的是，可能相同的子字符串被求解了很多遍，特别是.*对应的字符串需要循环求解。写完这个解法以后，我很自然地就知道可以用DP递归求解了，不过

原题：Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).For example,S = "ADOBECODEBANC"T = "ABC"Minimum window is "BANC".Note:

原题：原题：Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1...n.For example,Given n = 3, your program should return all 5 unique BST's shown below. 1

突然就写到72题了…中间也遇到了一些难题，60~70的题目有好几题坑题，不过Unique Path里有一题挺不错的，我想用迷宫回溯的方法求解，发现时间越界，其实也是用DP，有空放上来。 原题：Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (e

原题：Given an unsorted integer array, find the first missing positive integer.For example,Given [1,2,0] return 3,and [3,4,-1,1] return 2.Your algorithm should run in O(n) time and uses constant space.解

原题：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 only one unique solution.解决一个数独，我认为还是比较直观的，一直觉得计算

原题：Given an array of non-negative integers, you are initially positioned at the first index of the array.Each element in the array represents your maximum jump length at that position.Your goal is to r

又是一道Hard题，但AC却很简单，用暴力算法，大概5-10分钟就搞定了，比较难的是这题有很多种解法，把这些解法都理解一遍比较好。四种解法都做了注释，可以直接看代码。/*原题：Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) p

很有意思的问题，也让我重新看了一下第一题。 原题：Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.Note: The solution s

原题：Given a digit string, return all possible letter combinations that the number could represent.A mapping of digit to letters (just like on the telephone buttons) is given below.（一个手机的九宫格）虽然这题不是很难，但是我

做到这题的时候，开始觉得不是很难，毕竟O(m+n)的遍历算法是比较容易的，题目要的log(m+n)的算法，乍一看也是很容易想到的，只要我们用QuickSort或者说二分法的想法找到这个rank=(m+n)/2的点就可以了。但是做下去的时候，发现边界条件太复杂了，很难理清楚，特别是奇数和偶数带来的0.5的偏差，让二分做不到真正的二分。相信看到这篇博客的同学也是刷过题，知道其中的难度的，因此本文就主要简

原题： Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find t

原题：Implement strStr().Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.刚看到这题，我就放了个string.find()上去：return haystack.find(needle);完美AC。 但是转念一想，这题虽然

原题：Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.For example,Given:s1 = "aabcc",s2 = "dbbca",When s3 = "aadbbcbcac", return true.When s3 = "aadbbbaccc", return false.