动态规划 Leetcode 139 单词拆分

单词拆分

Leetcode 139

学习记录自代码随想录

要点在注释中详细说明，注意迭代公式中是dpi不是dpj-i

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdlib.h>


bool wordBreak(char* s, char** wordDict, int wordDictSize) {

    // 1.dp[j]代表背包容量为字符串s的长度j时是否可以利用字典中单词拼接为字符串s，dp[j]为true和false，
    bool dp[strlen(s)+1];
    memset(dp, false, sizeof(dp));
    // 2.递推公式，if(j到i之间的字符串在wordDict中 && dp[j-i]为true) 则dp[j]为true
    // 3.dp数组初始化，dp[0] = true;
    dp[0] = true;
    // 4.遍历顺序，本题中字符串组合顺序不能改变所以实际上是排列问题，所以先遍历背包，再遍历物品
    for(int j = 0; j < strlen(s)+1; j++){
        for(int i = 0; i < j; i++){
            int length = j - i;
            char sub_str[length+1];
            strncpy(sub_str, s+i, length);
            sub_str[length] = '\0';
            for(int k = 0; k < wordDictSize; k++){
                if((strcmp(wordDict[k], sub_str) == 0) && dp[i]){
                    dp[j] = true;
                }
            }
            
        }
    }
    // 5.举例推导dp数组
    return dp[strlen(s)];
}

int main(){
    int n;
    scanf("%d", &n);
    while(n){
        char s[301];
        scanf("%s", &s);
        int wordDictSize;
        scanf("%d", &wordDictSize);
        char** wordDict = (char**)malloc((wordDictSize) * sizeof(char*));
        for(int i = 0; i < wordDictSize; i++){
            wordDict[i] = (char*)malloc(21 * sizeof(char));
            scanf("%s", wordDict[i]);
        }
        bool result = wordBreak(s, wordDict, wordDictSize);
        printf("%d", result);
        for(int i = 0; i < wordDictSize; i++){
            free(wordDict[i]);
        }
        free(wordDict);
    }
    return 0;
}