leetcode 27

27. Remove Element

Given an array and a value, 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 in place with constant memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:
Given input array nums = [3,2,2,3]val = 3

Your function should return length = 2, with the first two elements of nums being 2.

去除数列中指定的数,返回剩余数的个数n,并将剩余的数存入原数列的前n位。

代码如下:

 1 class Solution {
 2 public:
 3     int removeElement(vector<int>& nums, int val) {
 4         if(nums.empty())
 5         {
 6             return 0;
 7         }
 8         int n = nums.size();
 9         if(n == 1 && nums[0] == val)
10         {
11             return 0;
12         }
13         if(n == 1)
14         {
15             return 1;
16         }
17         int sum = 0;
18         int j = 0;
19         for(int i =0; i < n; i++)
20         {
21             if(nums[i] != val)
22             {
23                 nums[j] = nums[i];
24                 j++;
25                 sum++;
26             }
27         }
28         return sum;
29     }
30 };

哈哈, 又一次100%。

原文地址:https://www.cnblogs.com/shellfishsplace/p/5848576.html