0027._Remove_Element

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027. Remove Element

难度: Easy

刷题内容

原题连接

内容描述

Given an array nums and a value val, 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 1:

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

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

 It doesn't matter what you leave beyond the returned length.

Example 2:

 Given nums = [0,1,2,2,3,0,4,2], val = 2,

 Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

 Note that the order of those five elements can be arbitrary.

 It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

 // nums is passed in by reference. (i.e., without making a copy)
 int len = removeElement(nums, val);

 // any modification to nums in your function would be known by the caller.
 // using the length returned by your function, it prints the first len elements.
 for (int i = 0; i < len; i++) {
     print(nums[i]);
 }

解题方案

思路 1
**- 时间复杂度: O(N)**- 空间复杂度: O(1)**

保留两个指针 i 和 j，其中 i 是慢指针，j 是快指针。当 nums[j] 与给定的值相等时，递增 j 以跳过该元素。只要 nums[j] !== val,我们就复制 nums[j] 到 nums[i] 并同时递增两个索引。重复这一过程，直到 j 到达数组的末尾，该数组的新长度为 i。

代码：

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/**
 * @param {number[]} nums
 * @param {number} val
 * @return {number}
 */
var removeElement = function (nums,val) {
    let j = 0;
    for(let i = 0,len = nums.length; i<len; i++){
        if(nums[i] !== val){
            nums[j] = nums[i];
            j++
        }
    }
    return j;
};