[LeetCode] Word Break

[LeetCode] Word Break

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".

解題思路：

解法1，回溯法。但會超時。其時間復雜度為len!

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;
    }
};

d[i] = true, 如果s[0,...i]在字典裡面

d[i] = true, 如果存在k，0

d[i] = false, 不存在這樣的k

代碼如下：

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時間復雜度為O(len^2)
 
 
另外，一個非常好的解法就是添加一個字符在s前面，使得代碼更加簡潔。
 
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