leetcode轮回计划20181027

在本期的轮回计划中遇到了第一道最短路问题127，用的是SPFA来解

1. 121 Best Time to Buy and Sell Stock
   题意：允许买卖一次
2. 122 Best Time to Buy and Sell Stock II
   题意：允许买卖任意次
3. 125 Valid Palindrome
   题意：判断给定字符串是不是回文，不考虑空格和标点，忽略大小写
4. 127 Word Ladder
   题意：给定一个vector作为单词字典，给定一个开始单词和结束单词，返回从开始单词到结束单词的最短路径长度。
   思路：涉及到图的最短路，答案：

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector& wordList) {
        
        int size = wordList.size();
        vector> edge(size + 2,vector());
        bool flag = false;
        
        if(diff(beginWord,endWord) == 1) return 2;
        
        for(int i = 0;i < size;++ i){
            for(int j = i;j < size;++ j){
                if(diff(wordList[i],wordList[j]) == 1){
                    edge[i].push_back(j);
                    edge[j].push_back(i);
                }
                //beginWord : size ; endWord : size + 1
                if(diff(wordList[i],beginWord) == 1){
                    edge[size].push_back(i);
                }
                int toEnd = diff(wordList[i],endWord);
                // cout< que;
        que.push(size);
        vector dis(size + 2,(1LL << 31) - 1);
        dis[size] = 1;
        for(;!que.empty();que.pop()){
            int n = que.front();
            int d = dis[n];
            //松弛
            for(auto point : edge[n]){
                if(dis[point] > d + 1){
                    dis[point] = d + 1;
                    que.push(point);
                    if(point == size + 1) return dis[point];
                }
            }
        }
        return 0;
    }
    int diff(string & begin,string &end){
        int ret = 0;
        int i = 0;
        while(i < begin.size()){
            if(begin[i] != end[i]) ret ++;
            i ++;
        }
        return ret;
    }
};

5. 129 Sum Root to Leaf Numbers
   题意：使用每条根节点到叶节点的路径表示一个数字，给出这些数字的和
   思路：分治
6. 130 Surrounded Regions
   题意：将不与边界相连接的区域覆盖
   思路：queue