27. Remove Element

https://leetcode.com/problems/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.

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]);
}

Thoughts

移除给定数组中和val相等的数。把不是val的移到前面，双指针，p指向下个可放的位置，i指向当前待移的元素。

Code

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        p = 0
        for i, num in enumerate(nums):
            if num == val: continue
            nums[p] = num
            p += 1
        return p

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

Analysis

时间复杂度O(n), O(1)空间。

也可以左右反向指针做，但太麻烦了。