[Leetcode 1051] Height Checker

原题说明

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

Return the minimum number of students that must move in order for all students to be standing in non-decreasing order of height.

Notice that when a group of students is selected they can reorder in any possible way between themselves and the non selected students remain on their seats.

 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

解题思路

这道题可以有很多做法。最高效的方法是：
建立一个数组height_to_num, 其index为height，对应的值为height出现的次数，该数组可替代排序
遍历一遍heights数组来更新height_to_num。
然后再遍历一遍heights数组，同时从height_to_num中找到排序后的高度sort_height,
比较height和sort_height, 如果不同则返回结果+1。

示例代码 (cpp)

class Solution {
public:
    // Use height as vector index to aovid sorting
    int heightChecker(vector<int>& heights) {
        vector<int> height_to_num = vector<int>(101, 0);
        for (const auto height : heights) {
            height_to_num[height]++;
        }
        int switch_count = 0;
        int sort_height = 0;
        for (const auto height : heights) {
            while (height_to_num[sort_height] == 0) {
                ++sort_height;
            }
            if (sort_height != height) {
                ++switch_count;
            }
            height_to_num[sort_height]--;
        }
        return switch_count;
    }
    // use sort 
    // O(nlogn)
    int heightChecker(vector<int>& heights) {
        const auto original_heights = heights;
        sort(heights.begin(), heights.end());
        int switch_count = 0;
        for (int i = 0; i < heights.size(); ++i) {
            if (heights[i] != original_heights[i]) {
                ++switch_count;
            }
        }
        return switch_count;
    } 
    // use priority queue 
    // O(nlogn)
    int heightChecker(vector<int>& heights) {
        priority_queue<int, vector<int>, greater<int>> pq(heights.begin(), heights.end());
        int switch_count = 0;
        for (const auto height : heights) {
            int sort_height = pq.top();
            pq.pop();
            if (sort_height != height) {
                ++switch_count;
            }
        }
        return switch_count;
    } 
};

示例代码 (java)

class Solution {
    public int heightChecker(int[] heights) {
        int[] heightToFreq = new int[101];
        
        for (int height : heights) {
            heightToFreq[height]++;
        }
        
        int result = 0;
        int curHeight = 0;
        
        for (int i = 0; i < heights.length; i++) {
            while (heightToFreq[curHeight] == 0) {
                curHeight++;
            }
            
            if (curHeight != heights[i]) {
                result++;
            }
            heightToFreq[curHeight]--;
        }
        
        return result;
    }
}

示例代码 (python)

class Solution:
    def heightChecker(self, heights: List[int]) -> int:
        return sum(h1 != h2 for h1, h2 in zip(heights, sorted(heights)))

复杂度分析

时间复杂度: O(N)
空间复杂度: O(height)

归纳总结

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