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58. 最后一个单词的长度 #74

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Geekhyt opened this issue Sep 19, 2021 · 0 comments

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简单

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Geekhyt commented Sep 19, 2021

反向遍历

过滤掉末尾空格后，反向遍历字符串，并使用 count 计数，再次遇到空格时结束。

const lengthOfLastWord = function(s) {
    if (s.length === 0) return 0
    let count = 0
    for (let i = s.length - 1; i >= 0; i--) {
        if (s.charAt(i) === ' ') {
            if (count === 0) continue
            break
        }
        count++
    }
    return count
}

时间复杂度：O(n)
空间复杂度：O(1)

Geekhyt added the 简单 label

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