LeetCode097 Interleaving String

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详细见：leetcode.com/problems/interleaving-string

Java Solution: github

package leetcode;/* * 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. */public class P097_InterleavingString {/* * 已有的测试数据都通过，提交TLE */static class Solution {int count = 0;    public boolean isInterleave(String s1, String s2, String s3) {    if (s3 == null) {    return null == s2 && s1 == null;    }    if (s1 == null || s1.length() == 0) {    return s3.equals(s2);    } else if (s2 == null || s2.length() == 0) {    return s3.equals(s1);    }    if (s1.length() + s2.length() != s3.length()) {    return false;    }    boolean ans = solve(s1.toCharArray(), 0, s2.toCharArray(), 0, s3.toCharArray(), 0);        System.out.println(count);    return ans;    }private boolean solve(char[] ch1, int i1, char[] ch2, int i2, char[] ch3, int i3) {count ++;if (i1 == ch1.length && i3 < ch3.length) {return solve(ch2, i2, ch3, i3);} else if (i2 == ch2.length && i3 < ch3.length) {return solve(ch1, i1, ch3, i3);} else if (i3 == ch3.length) {return i1 == ch1.length && i2 == ch2.length;}if (ch1[i1] == ch2[i2]) {if (ch1[i1] != ch3[i3]) {return false;} else {return solve(ch1, i1 + 1, ch2, i2, ch3, i3 + 1) ||solve(ch1, i1, ch2, i2 + 1, ch3, i3 + 1);}}if (ch1[i1] == ch3[i3]) {return solve(ch1, i1 + 1, ch2, i2, ch3, i3 + 1);}if (ch2[i2] == ch3[i3]) {return solve(ch1, i1, ch2, i2 + 1, ch3, i3 + 1);}return false;}private boolean solve(char[] ch1, int i1, char[] ch2, int i2) {int j1 = i1, j2 = i2;while (j1 < ch1.length && j2 < ch2.length) {if (ch1[j1 ++] != ch2[j2 ++]) {return false;}}return j1 == ch1.length && j2 == ch2.length;}}/* *对于这类问题，肯定要尝试DP *12 ms */static class Solution2 {public boolean isInterleave(String s1, String s2, String s3) {if (s3 == null) {    return null == s2 && s1 == null;    }    if (s1 == null || s1.length() == 0) {    return s3.equals(s2);    } else if (s2 == null || s2.length() == 0) {    return s3.equals(s1);    }    int m = s1.length(), n = s2.length();    if (m + n != s3.length()) {    return false;    }    boolean[][] path = new boolean[m + 1][n + 1];    path[0][0] = true;    for (int i = 1; i < m + 1; i ++) {    path[i][0] = path[i - 1][0] & (s1.charAt(i - 1) == s3.charAt(i - 1));    }    for (int j = 1; j < n + 1; j ++) {    path[0][j] = path[0][j - 1] & (s2.charAt(j - 1) == s3.charAt(j - 1));    }    for (int i = 1; i < m + 1; i ++) {    for (int j = 1; j < n + 1; j ++) {    path[i][j] = (path[i - 1][j] & (s1.charAt(i - 1) == s3.charAt(i + j - 1))) ||    (path[i][j - 1] & (s2.charAt(j - 1) == s3.charAt(i + j - 1)));    }    }    return path[m][n];}}}

C Solution: github

/*    url: leetcode.com/problems/interleaving-string    recur: TLE    dp/dp2: AC 3ms 42.86%*/#include <stdio.h>#include <stdlib.h>#include <string.h>#define bool intbool cmp_str(char* s1, int i1, int n1, char* s2, int i2, int n2) {    if (n1-i1 != n2-i2) return 0;    while (i1 < n1) {        if (s1[i1] != s2[i2]) return 0;        i1 ++;        i2 ++;    }    return 1;}bool search(char* s1, int i1, int n1, char* s2, int i2, int n2, char* s3, int i3, int n3) {    if (i1 == n1) return cmp_str(s2, i2, n2, s3, i3, n3);    if (i2 == n2) return cmp_str(s1, i1, n1, s3, i3, n3);    if (s1[i1] == s3[i3] && search(s1, i1+1, n1, s2, i2, n2, s3, i3+1, n3))        return 1;    if (s2[i2] == s3[i3] && search(s1, i1, n1, s2, i2+1, n2, s3, i3+1, n3))        return 1;    return 0;}bool isInterleave_recur(char* s1, char* s2, char* s3) {    int n1 = s1 == NULL ? 0 : strlen(s1);    int n2 = s2 == NULL ? 0 : strlen(s2);    int n3 = s3 == NULL ? 0 : strlen(s3);    if (n1+n2 != n3) return 0;    return search(s1, 0, n1, s2, 0, n2, s3, 0, n3);}bool isInterleave_dp(char* s1, char* s2, char* s3) {    int n1 = s1 == NULL ? 0 : strlen(s1);    int n2 = s2 == NULL ? 0 : strlen(s2);    int n3 = s3 == NULL ? 0 : strlen(s3);    int i1 = 0, i2 = 0, ans = 0;    char** m = NULL;    if (n1+n2 != n3) return 0;    m = (char**) malloc(sizeof(char*) * (n1+1));    for (i1 = 0; i1 <= n1; i1 ++) {        m[i1] = (char*) malloc(sizeof(char) * (n2 + 1));        memset(m[i1], '0', n2+1);    }    m[0][0] = '1';    for (i1 = 1; i1 <= n1; i1 ++) {        if (s1[i1-1] != s3[i1-1]) break;        m[i1][0] = '1';    }    for (i2 = 1; i2 <= n2; i2 ++) {        if (s2[i2-1] != s3[i2-1]) break;        m[0][i2] = '1';    }    for (i1 = 1; i1 <= n1; i1 ++) {        for (i2 = 1; i2 <= n2; i2 ++) {            if (s1[i1-1] == s3[i1+i2-1]) {                if (m[i1-1][i2] == '1')                    m[i1][i2] = '1';            }            if (s2[i2-1] == s3[i1+i2-1]) {                if (m[i1][i2-1] == '1')                    m[i1][i2] = '1';            }        }    }    ans = m[n1][n2] == '1';    for (i1 = 0; i1 <= n1; i1 ++) free(m[i1]);    return ans;}bool isInterleave_dp2(char* s1, char* s2, char* s3) {    int n1 = s1 == NULL ? 0 : strlen(s1);    int n2 = s2 == NULL ? 0 : strlen(s2);    int n3 = s3 == NULL ? 0 : strlen(s3);    int i1 = 0, i2 = 0, ans = 0;    char* m = NULL;    if (n1+n2 != n3) return 0;    m = (char*) malloc(sizeof(char) * (n2 + 2));    memset(m, '0', n2+2);    m[n2+1] = '\0';    m[0] = '1';    for (i2 = 1; i2 <= n2; i2 ++) {        if (s2[i2-1] != s3[i2-1]) break;        m[i2] = '1';    }    for (i1 = 1; i1 <= n1; i1 ++) {        if (m[0] == '1' && s1[i1-1] != s3[i1-1])            m[0] = '0';        for (i2 = 1; i2 <= n2; i2 ++) {            if (s2[i2-1] == s3[i1+i2-1] && m[i2-1] == '1') {                m[i2] = '1';            } else if (s1[i1-1] == s3[i1+i2-1] && m[i2] == '1') {                m[i2] = '1';            } else  m[i2] = '0';        }    }    ans = m[n2] == '1';    free(m);    return ans;}bool isInterleave(char* s1, char* s2, char* s3) {    return isInterleave_dp2(s1, s2, s3);}int main() {    char* s1 = "aacaac";    char* s2 = "aacaaeaac";    char* s3 = "aacaaeaaeaacaac";    printf("answer is %d\r\n", isInterleave_dp(s1, s2, s3));    printf("answer is %d\r\n", isInterleave_recur(s1, s2, s3));        printf("answer is %d\r\n", isInterleave_dp2(s1, s2, s3));    return 0;}

Python Solution: github

#coding=utf-8'''    url: leetcode.com/problems/interleaving-string    @author:     zxwtry    @email:      zxwtry@qq.com    @date:       2017年4月25日    @details:    Solution: 92ms 18.64%'''class Solution(object):    def isInterleave(self, s1, s2, s3):        """        :type s1: str        :type s2: str        :type s3: str        :rtype: bool        """        n1 = 0 if s1 == None else len(s1)        n2 = 0 if s2 == None else len(s2)        n3 = 0 if s3 == None else len(s3)        if n1+n2 != n3: return False        dp = [[False for j in range(n2+1)] for i in range(n1+1)]        dp[0][0] = True        for j in range(1, n2+1):            if n2[j-1] != s3[j-1]: break            dp[0][j] = True        for i in range(1, n1+1):            if n1[i-1] != s3[i-1]: break            dp[i][0] = True        for i in range(1, n1+1):            for j in range(1, n2+1):                if s1[i-1] == s3[i+j-1] and dp[i-1][j]:                    dp[i][j] = True                if s2[j-1] == s3[i+j-1] and dp[i][j-1]:                    dp[i][j] = True        return dp[n1][n2]

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