(M)Dynamic Programming:139. Word Break
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建一个数组dp,dp[i]表示0~i是否可以被分成字典里的单词。遍历0到i的每个位置j,若0~j可以被分割成单词,那么只需要知道j~i是否在字典里就可以了。
class Solution {public: bool wordBreak(string s, vector<string>& wordDict) { int n = wordDict.size(); int len = s.size(); vector<bool> dp(len + 1, false); dp[0] = true; for(int i = 1; i <= len; ++i) { for(int j = 0; j <= i; ++j) { if(dp[j] && find(wordDict.begin(), wordDict.end(), s.substr(j, i - j)) != wordDict.end()) { dp[i] = true; break; } } } return dp.back(); }};
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