[每日 LeetCode] 27. Remove Element

Description:

Given an array nums and a value val, remove all instances of that value [ in-place ] (https://en.wikipedia.org/wiki/In-place_algorithm) 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 0,1,3,0,4.

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]);
}
​

思路：本题要求删除数组中为特定数值的元素，并返回最后的数组大小。遍历数组，将不等于 val 的元素依次放到数组的前面，最后调整新数组的长度。

C++ 代码

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int count = 0;
        for (int i=0; i<nums.size(); i++){
            if (nums[i] != val)
                nums[count++] = nums[i];
        }
        nums.resize(count);
        return count;
    }
};
​

运行时间：4ms

运行内存：8.4M