题目描述:
Given an array nums and a value val, remove all instances of that value in-placeand 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.
思路:
定义区间 [0,k)范围为无val的数列,遍历nums
代码如下:
class Solution {
public:
int removeElement(vector<int>& nums, int val) {
if(nums.size() == 0) return 0;
int k = 0;
for(int i = 0; i < nums.size(); ++i)
{
if(nums[i] != val)
nums[k++] = nums[i];
}
return k;
}
};