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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".
基本思路：
动态规划。
数组 dp[i]，表示第i个字符以前是否可以分隔成单词。 i 从0开始。
已知dp[0..i]时，求dp[i+1]，则需要偿试，s[k..i], 0<=k <=i, 进行偿试。
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在leetcode上实际执行时间为12ms。
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class Solution {
public:
bool wordBreak(string s, unordered_set
& wordDict) {
vector
dp(s.size()+1); dp[0] = true; for (int i=1; i<=s.size(); i++) { for (int j=0; j
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