27. 移除元素 Remove Element

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.

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 a 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]);
}

Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2]
Explanation: 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. For example if you return 2 with nums = [2,2,3,3] or nums = [2,3,0,0], your answer will be accepted.

 

方法一、双指针法

public int removeElement(int[] nums, int val) {
        int first = 0, second = 0;
        for (int i = 0; i < nums.length; i++){
            if(nums[second] == val) second++;
            else nums[first++] = nums[second++];
        }
        return first;
    }

 

方法二：

可以将删除元素与最后一个元素交换。

public int removeElement(int[] nums, int val) {
    int i = 0;
    int n = nums.length;
    while (i < n) {
        if (nums[i] == val) {
            nums[i] = nums[n - 1];
            // reduce array size by one
            n--;
        } else {
            i++;
        }
    }
    return n;
}

 

参考链接：

https://leetcode.com/problems/remove-element/

https://leetcode-cn.com/problems/remove-element/