1051. Height Checker
作者:互联网
问题:
给定一组学生身高,要对其进行身高排序,从低到高。
求最少要移动几个学生的位置。
Example 1: Input: heights = [1,1,4,2,1,3] Output: 3 Explanation: Current array : [1,1,4,2,1,3] Target array : [1,1,1,2,3,4] On index 2 (0-based) we have 4 vs 1 so we have to move this student. On index 4 (0-based) we have 1 vs 3 so we have to move this student. On index 5 (0-based) we have 3 vs 4 so we have to move this student. Example 2: Input: heights = [5,1,2,3,4] Output: 5 Example 3: Input: heights = [1,2,3,4,5] Output: 0 Constraints: 1 <= heights.length <= 100 1 <= heights[i] <= 100
解法:
基本解法:
将身高数组排序,得到最终排序结果,和原数组比较,同一位置i上有不一样的元素,res++
代码参考:
1 class Solution { 2 public: 3 int heightChecker(vector<int>& heights) { 4 int res=0; 5 vector<int>heightsOrd(heights); 6 sort(heightsOrd.begin(), heightsOrd.end()); 7 for(int i=0; i<heights.size(); i++){ 8 if(heights[i]!=heightsOrd[i]) res++; 9 } 10 return res; 11 } 12 };
不排序解法:
由于本问题的限制:身高在1~100,因此,我们可以使用桶排序的思想
设定高度计数数数组 htcnt[101]
然后再对比同一偏移,
htcnt的偏移:curh++ ,htcnt[curh]--
原高度数组偏移:h : heights
是否高度一样,不一样res++
代码参考:
1 class Solution { 2 public: 3 int heightChecker(vector<int>& heights) { 4 int htcnt[101]={0}; 5 for(int h:heights){ 6 htcnt[h]++; 7 } 8 int curh=0; 9 int res=0; 10 for(int h:heights){ 11 while(htcnt[curh]==0){ 12 curh++; 13 } 14 if(h!=curh){ 15 res++; 16 } 17 htcnt[curh]--; 18 } 19 return res; 20 } 21 };
标签:htcnt,1051,int,res,heights,++,Checker,Height,curh 来源: https://www.cnblogs.com/habibah-chang/p/13124440.html