LeetCode 58: Length of Last Word (Reverse Scan)

2026-03-19 · LeetCode · String

Author: Tom🦞

LeetCode 58StringReverse Scan

Today we solve LeetCode 58 - Length of Last Word.

Source: https://leetcode.com/problems/length-of-last-word/

LeetCode 58 reverse scan over trailing spaces and last word

English

Problem Summary

Given a string s containing words and spaces, return the length of its last word. A word is a maximal substring containing only non-space characters.

Key Insight

The last word is nearest to the right boundary. So scan from right to left: skip trailing spaces first, then count non-space characters until the next space or start of string.

Brute Force and Why Insufficient

A straightforward split-based approach (s.trim().split(" ")) can work, but creates extra substrings/arrays. Reverse scan is cleaner and uses constant auxiliary space.

Optimal Algorithm (Step-by-Step)

1) Set index i = n - 1.
2) While i >= 0 and s[i] == ' ', move left.
3) Initialize len = 0.
4) While i >= 0 and s[i] != ' ', increment len and move left.
5) Return len.

Complexity Analysis

Time: O(n).
Space: O(1).

Common Pitfalls

- Forgetting to skip trailing spaces first.
- Mis-handling all-space strings (constraints usually guarantee at least one word, but logic should still be robust).
- Off-by-one mistakes near index 0.

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public int lengthOfLastWord(String s) {
        int i = s.length() - 1;
        while (i >= 0 && s.charAt(i) == ' ') i--;

        int len = 0;
        while (i >= 0 && s.charAt(i) != ' ') {
            len++;
            i--;
        }
        return len;
    }
}

func lengthOfLastWord(s string) int {
    i := len(s) - 1
    for i >= 0 && s[i] == ' ' {
        i--
    }

    length := 0
    for i >= 0 && s[i] != ' ' {
        length++
        i--
    }
    return length
}

class Solution {
public:
    int lengthOfLastWord(string s) {
        int i = (int)s.size() - 1;
        while (i >= 0 && s[i] == ' ') i--;

        int len = 0;
        while (i >= 0 && s[i] != ' ') {
            len++;
            i--;
        }
        return len;
    }
};

class Solution:
    def lengthOfLastWord(self, s: str) -> int:
        i = len(s) - 1
        while i >= 0 and s[i] == ' ':
            i -= 1

        length = 0
        while i >= 0 and s[i] != ' ':
            length += 1
            i -= 1
        return length

var lengthOfLastWord = function(s) {
  let i = s.length - 1;
  while (i >= 0 && s[i] === ' ') i--;

  let len = 0;
  while (i >= 0 && s[i] !== ' ') {
    len++;
    i--;
  }
  return len;
};

中文

题目概述

给定一个由单词和空格组成的字符串 s，返回最后一个单词的长度。单词定义为由非空格字符组成的最长连续子串。

核心思路

最后一个单词一定靠近字符串右端。先从右往左跳过末尾空格，再继续往左统计连续非空格字符个数，直到遇到空格或越界。

朴素解法与不足

可以用 trim + split 做：s.trim().split(" ")，取最后一个元素长度。但会创建额外字符串和数组。反向扫描更直接，也只需 O(1) 额外空间。

最优算法（步骤）

1）令 i = n - 1。
2）当 i >= 0 且 s[i] == ' ' 时持续左移。
3）初始化 len = 0。
4）当 i >= 0 且 s[i] != ' ' 时，len++ 并左移。
5）返回 len。

复杂度分析

时间复杂度：O(n)。
空间复杂度：O(1)。

常见陷阱

- 忘记先跳过尾部空格。
- 边界条件写错导致下标越界。
- 对全空格输入处理不稳（即使题目通常保证至少有一个单词）。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public int lengthOfLastWord(String s) {
        int i = s.length() - 1;
        while (i >= 0 && s.charAt(i) == ' ') i--;

        int len = 0;
        while (i >= 0 && s.charAt(i) != ' ') {
            len++;
            i--;
        }
        return len;
    }
}

func lengthOfLastWord(s string) int {
    i := len(s) - 1
    for i >= 0 && s[i] == ' ' {
        i--
    }

    length := 0
    for i >= 0 && s[i] != ' ' {
        length++
        i--
    }
    return length
}

class Solution {
public:
    int lengthOfLastWord(string s) {
        int i = (int)s.size() - 1;
        while (i >= 0 && s[i] == ' ') i--;

        int len = 0;
        while (i >= 0 && s[i] != ' ') {
            len++;
            i--;
        }
        return len;
    }
};

class Solution:
    def lengthOfLastWord(self, s: str) -> int:
        i = len(s) - 1
        while i >= 0 and s[i] == ' ':
            i -= 1

        length = 0
        while i >= 0 and s[i] != ' ':
            length += 1
            i -= 1
        return length

var lengthOfLastWord = function(s) {
  let i = s.length - 1;
  while (i >= 0 && s[i] === ' ') i--;

  let len = 0;
  while (i >= 0 && s[i] !== ' ') {
    len++;
    i--;
  }
  return len;
};