leetcode 27 removeElement

最新推荐文章于 2025-08-17 23:50:11 发布

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最新推荐文章于 2025-08-17 23:50:11 发布

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本文介绍了一种在C++中高效移除数组内特定值的方法，并提供了三种不同的实现方式，重点在于如何不使用额外的空间就能完成这一操作。

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

Do not allocate extra space for another array, you must do this in place with constant memory.

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

Example:
Given input array nums = [3,2,2,3], val = 3

Your function should return length = 2, with the first two elements of nums being 2.

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        typedef vector<int>::iterator it;

        it i = nums.begin();
        it j = nums.end();

        while (i != j) {
            while (i != j && *i != val) ++i;

            if (i == j) break;

            --j;

            while (i != j && *j == val) --j;

            if (i != j) *i = *j;
        }

        nums.erase(j, nums.end());
        return nums.size();
    }
}; // runtime contribution 64.8%

hehe

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        nums.erase(remove(nums.begin(), nums.end(), val), nums.end());
        return nums.size(); 
    }
}; // runtime contribution 10.05%

参考后

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        typedef vector<int>::size_type sz;
        sz size = nums.size();

        for (sz i = 0; i < size;) {
            if (nums[i] == val) {
                nums[i] = nums[size - 1];
                --size;
            }else 
                ++i;
        }
        nums.erase(nums.begin() + size, nums.end());
        return size;

    }
}; // runtime contribution 64.8%