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2018-11-16.md的algorithm #2

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MarsForever opened this issue Nov 25, 2018 · 0 comments

MarsForever commented Nov 25, 2018 •

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第一个if文改成下面这种形式，避免了最后一个字母和target是同值的情况

//如果 target大于等于数组中最大的字母大，那么返回第一个字母
if (target >= letters[len-1]) {
       return letters[0];
}

那么下面这一步可以省略

if (mid + 1 == len) {
       return letters[0];
 }

大概可以快1ms左右

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