【记忆回溯】【深度搜索】【动态规划】【字符串】【力扣】单词拆分
目录
题目描述
解题思路
解答(c语言)
题目描述
给你一个字符串 s
和一个字符串列表 wordDict
作为字典。如果可以利用字典中出现的一个或多个单词拼接出 s
则返回 true
。
注意:不要求字典中出现的单词全部都使用,并且字典中的单词可以重复使用。
示例 1:
输入: s = "leetcode", wordDict = ["leet", "code"] 输出: true 解释: 返回 true 因为 "leetcode" 可以由 "leet" 和 "code" 拼接成。
示例 2:
输入: s = "applepenapple", wordDict = ["apple", "pen"] 输出: true 解释: 返回 true 因为 "applepenapple" 可以由 "apple" "pen" "apple" 拼接成。注意,你可以重复使用字典中的单词。
示例 3:
输入: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"] 输出: false
解题思路
解题思路:深度遍历的思想
从字符串的起始位置开始匹配字典中的单词,如果能匹配到,则继续从字符串中该单词后面继续匹配字典中的单词,直到完全匹配。
- 比如s = "catsandog",wordDict = {"cats","dog","sand","and","cat"}
- index = 0时,对"catsandog"进行前缀匹配,可以匹配到"cats",也可以匹配到"cat"等,则需要记忆回溯,也就是深度遍历,先匹配到"cats"处理,然后更新index = 4
- index = 4时,对"andog"进行前缀匹配,匹配到"and",然后更新index = 7
- index = 7时,对"og"进行前缀匹配,没有一个能匹配,然后返回到index = 4时,寻找能否有匹配的单词,已经匹配过的不再匹配,比如"and"
- 如果index = 4时没有可匹配的,返回index = 0时的前缀匹配,寻找能否有匹配的单词
- 如果index = 4时有可匹配的,则继续向下匹配,像刚才那样
- 就这样,来来回回寻找(深度遍历),直到index = strlen(s),说明可以s能被wordDict拼接。如果index从0到1,到2,到strlen(s) - 1对字典所有匹配都尝试完了,仍然没有则说明不能被拼接
注意点:已经匹配过的不再匹配,提交后才发现,否则会超时,因为会有很多重复判断
解答(c语言)
#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>int isMatch = 0; // 标记能否被分割int Forward(int index, char* s, size_t sLen, char **wordDict, int wordDictSize, int *isHasJudge)
{// 一旦有拼接成功的方案,则不再找寻其它方案if (isMatch == 1) {return 1;}// 恰好拼接成功if (index == sLen) {isMatch = 1;return 1;}// 无法拼接成功if (index > sLen) {return -1;}// 判断过的不再判断if (isHasJudge[index] == 1) {return -1;}// 遍历全部单词判断for (int j = 0; j < wordDictSize; j++) {size_t len = strlen(wordDict[j]);if (strncmp(s + index, wordDict[j], len) == 0) {int ret = Forward(index + len, s, sLen, wordDict, wordDictSize, isHasJudge);if (ret == 1) {return 1;}}}// 已判断过的做好标记isHasJudge[index] = 1;// 如果没有一个单词能匹配,则返回负数return -1;
}bool wordBreak(char* s, char **wordDict, int wordDictSize)
{// 每个用例会更改全局变量,需要重新初始化isMatch = 0; size_t sLen = strlen(s);// 记录每个下标(0,1,2,..., sLen-1)是否已经和所有子串匹配过int isHasJudge[sLen]; memset(isHasJudge, 0, sizeof(isHasJudge));// 深度遍历寻找所有方案Forward(0, s, sLen, wordDict, wordDictSize, isHasJudge);return isMatch == 1;
}int main()
{int select = 3;if (select == 1) {char *s = "catsandog";char *wordDict[] = {"cats","dog","sand","and","cat"};wordBreak(s, wordDict, 5);} else if (select == 2) {char *s = "applepenapple";char *wordDict[] = {"apple","pen"};wordBreak(s, wordDict, 2);} else if (select == 3) {char *s = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab";char *wordDict[] = {"a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"};wordBreak(s, wordDict, 10);}return 0;
}
end