LeetCode:139.Word Break
来源：互联网 发布：男朋友礼物 知乎 编辑：程序博客网 时间：2024/06/05 09:54
Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.
For example, given
s = "leetcode",
dict = ["leet", "code"].
Return true because "leetcode" can be segmented as "leet code".
- Total Accepted: 132579
- Total Submissions: 460844
- Difficulty: Medium
- Contributors: Admin
题目的意思是，已给一个字符串数组dict，与一个字符串s，查看s是否能由数组dict中的一个或多个字符串组成。
一开始最快想到的是递归，代码如下：
class Solution {public:bool wordBreak(string s, vector<string>& wordDict) {if (s.size() == 0)return true;if (wordDict.size() == 0)return false;for (int i = 0;i < s.size();i++){if (find(wordDict.begin(), wordDict.end(), s.substr(0, i + 1)) != wordDict.end()){if (wordBreak(s.substr(i + 1, s.size() -1- i), wordDict))return true;}}return false;}};果然超时了。
一般递归超时的题目会想到动态规划的方法，的确这题字符串的每个状态都是以前面的有关的，例子：
s = "leetcode",
dict = ["leet", "code"]
倒着想，如果检测到字符c，如果在c之前的“leet”能够在数组里面成功表示，且c也是可以在数组中找到，那么“leetc”是满足条件的，只要满足一种情况即可。代码：
class Solution { public:bool wordBreak(string s, vector<string> &wordDict) { if (s.size()!=0 && wordDict.size()== 0) return false; bool *record = new bool[s.size()+1]; for (int i = 0;i<s.size()+1;i++) record[i] = false; record[0] = true; for (int i = 1; i < s.size() + 1; i++) { for (int j = i - 1; j >= 0; j--) { if (record[j] && find(wordDict.begin(),wordDict.end(),s.substr(j, i - j)) != wordDict.end()) { record[i] = true; break; //只要能满足一种情况即可 } } } return record[s.size()]; }};34 / 34 test cases passed.Status:Accepted
Runtime: 3 ms
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