Question

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]); }

Difficulty:Easy

Category:Array, Two Points

Analyze

这道题目只需要去除与目标数值相同的元素既可， 可以有以下几种方式去实现：

方案一： 直接跳过重复的元素， 时间复杂度O(n), 空间复杂度O(1) 方案二： 使用Remove函数，时间复杂度O(n), 空间复杂度O(1)

Solution

Solution 1(方案一)： 直接跳过目标位置

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

Solution 1(方案一)： 使用std::remove函数以及distance函数

class Solution {
 public:
  int removeElement(vector<int>& nums, int val) {
    return distance(nums.begin(), remove(nums.begin(), nums.end(), val));
  }
};