letecode [27] - 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.

题目大意：

　　给定一个无序数组，删除与指定元素相同的所有元素，返回操作后的数组长度。

理  解 ：

　　该题类似于上个删除重复元素的题目。方法也类似。

　　设置两个“指针” ，j 指针用来遍历比较判断是否满足条件， i 指针用来保存 新的数组元素（即未删除的元素）。

代 码 C++：

　　

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

 

运行结果：

　　执行用时 : 8 ms　　内存消耗 : 8.7 MB