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好爽！！其实这题dfs可以过，我的就是，16ms，最重要的是中间加一步剪枝就是如果当前string 1 的pos == string 2 待检测的pos (也就是上次检查的pos+1)，就一定匹配，直接pos++,检查下一个元素.. Followed by:
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