算法题丨Remove Element

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 by modifying the input array in-place with O(1) extra memory.
The order of elements can be changed. It doesn't matter what you leave beyond the new length.

示例 Given nums = [3,2,2,3], val = 3, Your function should return length = 2, with the first two elements of nums being 2. 算法分析

难度:低
分析:给定数组和指定一个目标值,从数组中移除所有跟目标值相等的元素,返回最终元素的长度,注意不要另外分配内存空间。
思路:题目很简单,直接遍历数组元素,判断当前元素是否跟目标值相等,如果不相等,证明当前元素应该留在数组中,有效数组长度自增1,否则为无效元素,因为只需返回有效数组长度,所以不用删除元素,跳过此循环即可。

代码示例(C#) public int RemoveElement(int[] nums, int val) { int i = 0; for (int j = 0; j < nums.Length; j++) { //如果不相等,有效长度自增1 if (nums[j] != val) { nums[i] = nums[j]; i++; } } return i; } 复杂度

时间复杂度:O (n).

空间复杂度:O (1).

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