zhidao_wenge的专栏

vectorstring> fullJustify(vectorstring> &words, int L) { vectorstring> result; if(words.empty() || L 0) return result; int i = 0, begin = 0, end = 0; int

vector tmp_result;     vector > result;     vector > is_division;     int str_length;     void dpDivision(const string &str)     {         for(int i = str_length - 1; i >= 0; --i)         {

int length;     vector one_result;     vector result;     vector > is_word;          void backTrack(string &s, int last, int index, vector &one_result)     {         if(last == length)        

vectorstring> one_result; vectorvectorstring> > result; mapstring, int> word_deep_map; //某个单词在第几层上。 int word_lenght; /* 宽度搜索 建立graph（记录单词的层次） */ void bfs(string &start, string &end, setstri

动态规划：假设抢到第i个房间了， money[i] = max(num[i] + money[i-2],  money[i-1]); 因为是一个圈，要记录i是来自哪里，用于判断最后一个的最大值是否来自最开始的那个（ffuckk，不会描述） vector money;     vector pre_index;//记录当前这个钱 是从哪个房间来的     int lenght;     

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 该方法的核心是将原问题转变成一个寻

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 该方法的核心是将原问题转变成一个寻