Leetcode解题报告：Remove Element

题目

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.

public class Solution {
    public int removeElement(int[] nums, int val) {
    }
}

Leetcode链接

https://leetcode.com/problems...

算法

解决这个问题需要定义两个指针，i和j，i指向值为val的数字，j指向值不为val的数字，然后将nums[i]的值设置为nums[j]即可，然后将i和j向数组后面移动，并在满足以上条件的时候再次对num[i]进行赋值，一直到我们遍历到数组末尾位置。

Java代码：

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

这种算法的时间复杂度为O(n)，空间复杂度O(1)。