README.md

Remove Element

Description

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.

Example:

Given nums = [3,2,2,3], val = 3,

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

Tags: Array, Two Pointers

思路

题意是移除数组中值等于 val 的元素，并返回之后数组的长度，并且题目中指定空间复杂度为 O(1)，我的思路是用 tail 标记尾部，遍历该数组时当索引元素不等于 val 时，tail 加一，尾部指向当前元素，最后返回 tail 即可。

class Solution {
    public int removeElement(int[] nums, int val) {
        int tail = 0;
        for (int i = 0, len = nums.length; i < len; ++i) {
            if (nums[i] != val) {
                nums[tail++] = nums[i];
            }
        }
        return tail;
    }
}

kotlin(220ms/92.31%):

class Solution {
    fun removeElement(nums: IntArray, `val`: Int): Int {
        var count = 0
        for (index in nums.indices) {
            val t = nums[index]
            if (t != `val`) {
                nums[count++] = t
            }
        }
        return count
    }
}

javascript

var removeElement = function(nums, val) {
    var len = nums.length
    for(var i = 0; i < len; i++) {
        if (nums[i] === val) {
            nums.splice(i, 1)
            i = i - 1
        }
    }
    return nums.length
};

结语

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