27. Remove Element #

题目 #

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]);
}

题目大意 #

给定一个数组 nums 和一个数值 val，将数组中所有等于 val 的元素删除，并返回剩余的元素个数。

解题思路 #

这道题和第 283 题很像。这道题和第 283 题基本一致，283 题是删除 0，这一题是给定的一个 val，实质是一样的。

这里数组的删除并不是真的删除，只是将删除的元素移动到数组后面的空间内，然后返回数组实际剩余的元素个数，OJ 最终判断题目的时候会读取数组剩余个数的元素进行输出。

代码 #

package leetcode

func removeElement(nums []int, val int) int {
	if len(nums) == 0 {
		return 0
	}
	j := 0
	for i := 0; i < len(nums); i++ {
		if nums[i] != val {
			if i != j {
				nums[i], nums[j] = nums[j], nums[i]
			}
			j++
		}
	}
	return j
}