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Word Break

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

解题思路：

解法1，回溯法。但会超时。其时间复杂度为len!

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class Solution {
public:
    bool wordBreak(string s, unordered_set
  
   & wordDict) {
        return helper(s, wordDict);
    }
    
    bool helper(string s, unordered_set
   
    & wordDict){ if(s==""){ return true; } int len = s.length(); for(int i=1; i<=len; i++){ if(wordDict.find(s.substr(0, i))!=wordDict.end() && helper(s.substr(i), wordDict)){ return true; } } return false; } };
   
  

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d[i] = true, 如果s[0,...i]在字典里面

d[i] = true, 如果存在k，0

d[i] = false, 不存在这样的k

代码如下：

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class Solution {
public:
    bool wordBreak(string s, unordered_set
  
   & wordDict) {
        int len = s.length();
        if(len == 0){
            return true;
        }
        bool d[len];
        memset(d, 0, len * sizeof(bool));
        if(wordDict.find(s.substr(0, 1)) == wordDict.end()){
            d[0] = false;
        }else{
            d[0] = true;
        }
        for(int i=1; i
   
    时间复杂度为O(len^2)
    
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另外，一个非常好的解法就是添加一个字符在s前面，使得代码更加简洁。
    
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class Solution {
public:
    bool wordBreak(string s, unordered_set
     
      & wordDict) {
        s = "#" + s;
        int len = s.length();
        bool d[len];
        memset(d, 0, len * sizeof(bool));
        d[0] = true;
        for(int i=1; i
      
       
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