lintcode interleaving-string 交叉字符串

问题描述

lintcode

笔记

buff[i][j]表示：
s1的前i个字符和s2的前j个字符是否能交叉组成s3前i+j个字符。
若要buff[i][j] = true,有两种情况。

• s1前i-1个字符与s2前j个字符符合要求，而且s1的i个字符正好是s3的第(i+j)个字符。
• s2前j-1个字符与s1前i个字符符合要求，而且s2的j个字符正好是s3的第(i+j)个字符。

还要做初始化。见代码2的再次练习

代码

class Solution {public:    /**     * Determine whether s3 is formed by interleaving of s1 and s2.     * @param s1, s2, s3: As description.     * @return: true of false.     */    bool isInterleave(string s1, string s2, string s3) {        // write your code here        const int len1 = s1.size();        const int len2 = s2.size();        const int len3 = s3.size();        if (len1 + len2 != len3)            return false;        vector<vector<bool> > dp(len1+1, vector<bool>(len2+1, false));        dp[0][0] = true;        for (int i = 1; i <= len1; i++)            dp[i][0] = dp[i-1][0] && (s1[i-1] == s3[i-1]);        for (int j = 1; j <= len2; j++)            dp[0][j] = dp[0][j-1] && (s2[j-1] == s3[j-1]);        for (int i = 1; i <= len1; i++)        {            for (int j = 1; j <= len2; j++)            {                int t = i + j;                if (s3[t-1] == s1[i-1])                    dp[i][j] = dp[i][j] || dp[i-1][j];                if (s3[t-1] == s2[j-1])                    dp[i][j] = dp[i][j] || dp[i][j-1];            }        }        return dp[len1][len2];    }};

再次练习

class Solution {public:    /**     * Determine whether s3 is formed by interleaving of s1 and s2.     * @param s1, s2, s3: As description.     * @return: true of false.     */    bool isInterleave(string s1, string s2, string s3) {        // write your code here        const int len1 = s1.size();        const int len2 = s2.size();        if (len1 + len2 != s3.size())            return false;        bool buff[len1+1][len2+1];        buff[0][0] = true; // 用s1的前0位（空字符串）与s2的前0位（空字符串），可以组成s3的前0位（空字符串）        for (int i = 1; i <= len1; i++)// 用s1的前i位，和s2的前0位（空字符串）        {            buff[i][0] = buff[i-1][0] && s1[i-1]==s3[i-1];        }        for (int j = 1; j <= len2; j++)// 用s1的前0位（空字符串），和s2的前j位        {               buff[0][j] = buff[0][j-1] && s2[j-1]==s3[j-1];        }        for (int i = 1; i <= len1; i++)        {            for (int j = 1; j <= len2; j++)            {                buff[i][j] = (buff[i-1][j] && s1[i-1]==s3[i+j-1]) || (buff[i][j-1] && s2[j-1]==s3[i+j-1]);                // cout << i << ' ' << j << ' ' << buff[i][j] << endl;            }        }        return buff[len1][len2];    }};