[LeetCode] 1051. Height Checker

Easy

Students are asked to stand in non-decreasing order of heights for an annual photo.

Return the minimum number of students not standing in the right positions.  (This is the number of students that must move in order for all students to be standing in non-decreasing order of height.)

 

Example 1:

Input: [1,1,4,2,1,3]
Output: 3
Explanation: 
Students with heights 4, 3 and the last 1 are not standing in the right positions.

 

Note:

1. 1 <= heights.length <= 100
2. 1 <= heights[i] <= 100

题目大意：给出一组表示学生身高的数字，学生身高应该非降序排列，求出已给的顺序中未按照要求排在自己位置上的数字的个数。

比如[1,1,4,2,1,3]，输出结果是3，因为这个顺序中4,3和最后一个1未排到要求的位置上。

 

这道题一个思路是对这个数组进行排序，然后比较排序前和排序后的数组对应元素不相等的个数。

代码如下：

class Solution {
public:
    int heightChecker(vector<int>& heights) {
        if (heights.empty())return 0;
        int res = 0;
        vector<int> temp=heights;
        for(int i=1;i<temp.size();++i){
            if(temp[i]<temp[i-1]){
                int m = i, n = i-1;
                while (n>=0) {
                    if (temp[m] < temp[n]) {
                        swap(temp[m], temp[n]);
                        m--;
                    }
                    n--;
                }
            }
        }
        for(int i=0;i<heights.size();++i){
            if(temp[i]!=heights[i])res++;
        }
        return res;
    }
};

这里我自己写了排序函数，也可以直接调用排序函数sort()

class Solution {
public:
    int heightChecker(vector<int>& heights) {
        if (heights.empty())return 0;
        int res = 0;
        vector<int> temp=heights;
        sort(temp.begin(),temp.end());
        for(int i=0;i<heights.size();++i){
            if(temp[i]!=heights[i])res++;
        }
        return res;
    }
};