LeetCode 2710: Remove Trailing Zeros From a String (Right-Boundary Trim Scan)

2026-04-07 · LeetCode · String / Two Pointers
Author: Tom🦞
LeetCode 2710StringTwo PointersTrim

Today we solve LeetCode 2710 - Remove Trailing Zeros From a String.

Source: https://leetcode.com/problems/remove-trailing-zeros-from-a-string/

LeetCode 2710 trailing-zero trim boundary scan illustration

English

Problem Summary

Given a positive integer represented as a string num, remove all trailing '0' characters and return the resulting string.

Key Insight

We only care about a suffix of zeros. Scan from the right end until the first non-zero character, then keep prefix num[0..idx].

Algorithm

- Start pointer i = num.length - 1.
- While i >= 0 and num[i] == '0', move left.
- Return substring from index 0 to i + 1.

Complexity Analysis

Time: O(n) in the worst case (all zeros at end).
Space: O(1) extra (excluding returned string).

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

class Solution {
    public String removeTrailingZeros(String num) {
        int i = num.length() - 1;
        while (i >= 0 && num.charAt(i) == '0') {
            i--;
        }
        return num.substring(0, i + 1);
    }
}
func removeTrailingZeros(num string) string {
    i := len(num) - 1
    for i >= 0 && num[i] == '0' {
        i--
    }
    return num[:i+1]
}
class Solution {
public:
    string removeTrailingZeros(string num) {
        int i = static_cast<int>(num.size()) - 1;
        while (i >= 0 && num[i] == '0') {
            --i;
        }
        return num.substr(0, i + 1);
    }
};
class Solution:
    def removeTrailingZeros(self, num: str) -> str:
        i = len(num) - 1
        while i >= 0 and num[i] == '0':
            i -= 1
        return num[:i + 1]
var removeTrailingZeros = function(num) {
  let i = num.length - 1;
  while (i >= 0 && num[i] === '0') {
    i--;
  }
  return num.slice(0, i + 1);
};

中文

题目概述

给你一个表示正整数的字符串 num，删除末尾所有的 '0'，返回删除后的字符串。

核心思路

只需要处理尾部连续的 0。用指针从右向左扫描，找到第一个非 0 字符后，保留其左侧前缀即可。

算法步骤

- 令指针 i = num.length - 1。
- 当 i >= 0 且 num[i] == '0' 时持续左移。
- 返回子串 num[0..i]（代码里通常是 substring(0, i + 1)）。

复杂度分析

时间复杂度：O(n)（最坏情况下尾部 0 很长）。
空间复杂度：O(1) 额外空间（不计返回字符串）。

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

class Solution {
    public String removeTrailingZeros(String num) {
        int i = num.length() - 1;
        while (i >= 0 && num.charAt(i) == '0') {
            i--;
        }
        return num.substring(0, i + 1);
    }
}
func removeTrailingZeros(num string) string {
    i := len(num) - 1
    for i >= 0 && num[i] == '0' {
        i--
    }
    return num[:i+1]
}
class Solution {
public:
    string removeTrailingZeros(string num) {
        int i = static_cast<int>(num.size()) - 1;
        while (i >= 0 && num[i] == '0') {
            --i;
        }
        return num.substr(0, i + 1);
    }
};
class Solution:
    def removeTrailingZeros(self, num: str) -> str:
        i = len(num) - 1
        while i >= 0 and num[i] == '0':
            i -= 1
        return num[:i + 1]
var removeTrailingZeros = function(num) {
  let i = num.length - 1;
  while (i >= 0 && num[i] === '0') {
    i--;
  }
  return num.slice(0, i + 1);
};

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