leetcode 27. 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 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.

Hint:

1. Try two pointers.
2. Did you use the property of "the order of elements can be changed"?
3. What happens when the elements to remove are rare?

删除数组中的val值,返回长度.

看到这题首先感觉和26题Remove Duplicates from Sorted Array一样,限制了思路.

public class A27RemoveElement {    public int removeElement(int[] nums, int val) {    if(nums == null || nums.length == 0) return 0;        int len = 0;        for(int i = 0; i < nums.length; i++) {        if(nums[i] != val) {        nums[len] = nums[i];        len++;        }        }        return len;    }}

注意到这句话"the order of elements can be changed",元素的顺序可以被改变.

看讨论区中有这样的解法,碰到val值.就用数组的最后一个元素来填补,索引不变,继续判断.这样减少了重复赋值.

public int removeElement(int[] nums, int val) {if(nums == null || nums.length == 0) return 0;    int len = 0;    int n = nums.length;    while (len < n) {        if (nums[len] == val) {            nums[len] = nums[n - 1];            n--;        } else {            len++;        }    }    return n;}