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".

解法：动态规划

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int n = s.length();
        int matrix[n][n];
        for(int i = 0;i<n;i++)
            memset(matrix[i],0,n*sizeof(int));
        for(int l = 1;l<=n;l++)
        {
            for(int i = 0;i<n-l+1;i++)
            {
                int j = i+l-1;
                if(dict.find(s.substr(i,l)) != dict.end()) {
                    matrix[i][j] = 1;
                    continue;
                }
                for(int k = i;k<j;k++)
                {
                    if(matrix[i][k] == 1 && matrix[k+1][j] == 1) matrix[i][j] = 1;
                }
            }
        }
        return (matrix[0][n-1] == 1)?true:false;  
    }
};

36 ms.

也可以写成

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int n = s.length();
        int dp[n+1];
        memset(dp,0,(n+1)*sizeof(int));
        dp[0] = 1;
        for(int i = 0;i<n;i++)
        {
            for(int j = i;j>=0;j--){
                if(dp[j] && dict.find(s.substr(j,i-j+1)) != dict.end()){
                    dp[i+1] = 1;
                    break;
                }
            }
        }
        return (dp[n]==1)?true:false;
    }
};

20 ms