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这样做不行啊m[i]记录匹配到第i个字母所需要的最小单词数 for j=0-〉i-1 for k=0-〉n(单词数) 如果words[k]==数字j+1-〉i位 找到最小的m[j]+1=m[i] 超时了 Followed by: Post your reply here: |
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