LeetCode 题解(229) : Remove Element

最新推荐文章于 2025-09-10 22:03:14 发布

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最新推荐文章于 2025-09-10 22:03:14 发布

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分类专栏：算法文章标签： algorithm leetcode 面试题

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本文介绍了一种C++方法来移除数组中的指定元素，并返回新数组的长度。通过双指针技巧实现原地操作，避免了额外的空间开销。

题目：

Given an array and a value, remove all instances of that value in place and return the new length.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

题解：

C++版：

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int i = 0, j = nums.size() - 1;
        while(i <= j) {
            while(i <= j && nums[i] != val)
                i++;
            while(i <= j && nums[j] == val)
                j--;
            if(i <= j) {
                nums[i] = nums[j];
                i++;
                j--;
            }
        } 
        return j + 1;
    }
};

Java版：

public class Solution {
    public int removeElement(int[] nums, int val) {
        int i = 0, j = nums.length - 1;
        while(i <= j) {
            while(i <= j && nums[i] != val)
                i++;
            while(i <= j && nums[j] == val)
                j--;
            if(i <= j) {
                nums[i] = nums[j];
                i++;
                j--;
            }
        } 
        return j + 1;
    }
}

Python版：

class Solution(object):
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        if len(nums) == 0:
            return 0
            
        i, j = 0, len(nums) - 1
        while i <= j:
            while i <= j and nums[i] != val:
                i += 1
            while j >= i and nums[j] == val:
                j -= 1
            if i <= j:
                nums[i] = nums[j]
                i += 1
                j -= 1
    
        return j + 1