[Leetcode] Remove Element

最新推荐文章于 2024-02-23 09:08:20 发布

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本文介绍了一种在数组中就地移除指定值的方法，并返回新的有效长度。该方法不使用额外空间，且时间复杂度较低。示例中给出了具体的实现代码及运行示例。

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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.

注意点：

1.交换两个值时的ID处理

2.避免不必要的循环

int removeElement(int *nums, int numsSize, int val) {
/*
	int i = 0;
	int j = 0;
	
	for(j = 0; j < numsSize; j ++){
		if (nums[j] != val){
			nums[i] = nums[j];
			i ++;
		}
	}
	
	return i;*/
	
	int i = 0; 
	int n = numsSize;
	
	while(i < n){
		if (nums[i] == val){
			nums[i] = nums[n - 1];
			n --;
		}
		else{
			i ++;
		}
	}
	
	return n;
}