[LeetCode] 1051. Height Checker

A school is trying to take an annual photo of all the students. The students are asked to stand in a single file line in non-decreasing order by height. Let this ordering be represented by the integer array expected where expected[i] is the expected height of the ith student in line.

You are given an integer array heights representing the current order that the students are standing in. Each heights[i] is the height of the ith student in line (0-indexed).

Return the number of indices where heights[i] != expected[i].

Example 1:
Input: heights = [1,1,4,2,1,3]
Output: 3
Explanation:
heights: [1,1,4,2,1,3]
expected: [1,1,1,2,3,4]
Indices 2, 4, and 5 do not match.

Example 2:
Input: heights = [5,1,2,3,4]
Output: 5
Explanation:
heights: [5,1,2,3,4]
expected: [1,2,3,4,5]
All indices do not match.

Example 3:
Input: heights = [1,2,3,4,5]
Output: 0
Explanation:
heights: [1,2,3,4,5]
expected: [1,2,3,4,5]
All indices match.

Constraints:
1 <= heights.length <= 100
1 <= heights[i] <= 100

高度检查器。

学校在拍年度纪念照时，一般要求学生按照 非递减 的高度顺序排列。

请你返回能让所有学生以 非递减 高度排列的最小必要移动人数。

注意，当一组学生被选中时，他们之间可以以任何可能的方式重新排序，而未被选中的学生应该保持不动。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/height-checker
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思路一 - 排序

因为题目问的是需要移动多少个人的位置，所以当排序过后，每个人就知道他们应该站在什么位置上了。此时把排序过的和未排序过的数组进行比较，有不同的，则说明需要移动。

复杂度

时间O(nlogn)
空间O(n)

代码

Java实现

class Solution {
    public int heightChecker(int[] heights) {
        int len = heights.length;
        int[] sorted = new int[len];
        for (int i = 0; i < len; i++) {
            sorted[i] = heights[i];
        }
        Arrays.sort(sorted);

        int count = 0;
        for (int i = 0; i < len; i++) {
            if (sorted[i] != heights[i]) {
                count++;
            }
        }
        return count;
    }
}

思路二 - 桶排序

桶排序的做法类似，但是更高效。因为身高的数据范围是 [1, 100]，所以我们创建 101 个桶，表示身高从 0 到 100 每个不同身高分别有多少人。第一次遍历数组，记录不同身高分别有多少人。再次遍历数组，因为需要满足排列是非递减，把桶中的元素按照桶的下标从小到大一个个放到 expected 数组内，然后再比较 expected 数组和 heights 数组，有不同的，则说明需要移动。

复杂度

时间O(n)
空间O(n)

代码

Java实现

class Solution {
    public int heightChecker(int[] heights) {
        int[] map = new int[101];
        for (int height : heights) {
            map[height]++;
        }

        int n = heights.length;
        int[] expected = new int[n];
        int index = 0;
        for (int i = 0; i < map.length; i++) {
            while (map[i] != 0) {
                expected[index++] = i;
                map[i]--;
            }
        }

        int diff = 0;
        for (int i = 0; i < n; i++) {
            if (expected[i] != heights[i]) {
                diff++;
            }
        }
        return diff;
    }
}