您的位置:首页 > 其它

LeetCode 97: Interleaving String 字符串交错

2015-09-01 22:50 411 查看
博客转载请注明地址:http://blog.csdn.net/SunliyMonkey/article/details/48165711

题目描述

题目地址:https://leetcode.com/problems/interleaving-string/

给定三个字符串S1S_1,S2S_2,S3S_3,判断S3S_3是否能够由S1S_1,S2S_2交错组合而成。

样例:

S1=aabccS_1 = aabcc

S2=dbbcaS_2 = dbbca

当S3=aadbbcbcacS_3 = aadbbcbcac时,返回 true

当S3=aadbbbacccS_3 = aadbbbaccc时,返回 false

考察点

动态规划

陷阱



Code

[code]class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
       vector <int> match[2];
        int cur, nxt;
        int l;
        int l1 = s1.length();
        int l2 = s2.length();
        int l3 = s3.length();

        if(l1+l2 != l3)
            return false;

        int *vis[2];

        vis[0] = new int[l1+1];
        vis[1] = new int[l1+1];

        for(int i = 0; i <= l1; i++)
            vis[0][i] = vis[1][i] = -1;

        nxt = 0;
        match[0].push_back(0);

        for(int i = 0; i < l3; i++){
            cur = nxt;
            nxt = !nxt;
            match[nxt].clear();

            for(int k = 0; k < match[cur].size(); k++){

                l = match[cur][k];
                if(l < l1 && s1[l] == s3[i]){

                    if(vis[nxt][l+1] != i){
                        match[nxt].push_back(l+1);
                        vis[nxt][l+1] = i;
                    }
                }

                if(i-l+1 <= l2 && s2[i-l] == s3[i]){

                    if(vis[nxt][l] != i){
                        match[nxt].push_back(l);
                        vis[nxt][l] = i;
                    }
                }
            }
        }
        return !match[nxt].empty();
    }
};


排名

内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: