题目

单个文件行

a single file line

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

解法

方法一：双重for循环（笨死了）

class Solution {
          
   
        public int heightChecker(int[] heights) {
          
   
            int[] unSortedHeights = new int[heights.length];


            for (int i = 0; i < heights.length; i++) {
          
   
                unSortedHeights[i] = heights[i];
            }

//            System.out.println("unSortedHeights -->" + unSortedHeights);

            Arrays.sort(heights);

//            System.out.println("heights --> " + heights);

            int sortedHeigthsLength = heights.length;
            int unSortedHeightsLength = unSortedHeights.length;

            int res = 0;

            for (int i = 0; i < unSortedHeightsLength; i++) {
          
   
                for (int j = 0; j < sortedHeigthsLength; j++) {
          
   
                    if (i == j && unSortedHeights[i] != heights[j]) {
          
   
//                        System.out.println("unSortedHeights[i] -- > " + unSortedHeights[i]);
//                        System.out.println("heights[j] -- > " + heights[j]);
                        res += 1;
                    }
                }
            }

            return res;
        }
    }

方法三： 基于比较的排序

代码

static class Solution2 {
          
   
        public static int heightChecker(int[] heights) {
          
   
            int l = heights.length, ans = 0;
            int[] expected = new int[l];

            System.arraycopy(heights, 0, expected, 0, l);
            Arrays.sort(expected);

            for (int i = 0; i < l; i++) {
          
   
                if (heights[i] == expected[i]) {
          
   
                    ans++;
                }
            }

            return ans;
        }
    }

复杂度分析

时间复杂度：O(nlogn)，其中 n 是数组 heights 的长度。即为排序需要的时间。 空间复杂度：O(n)，即为数组 expected 需要的空间。

方法三：计数排序

代码

static class Solution3 {
          
   
        public static int heightChecker(int[] heights) {
          
   
            int m = Arrays.stream(heights).max().getAsInt();

            System.out.println(m);

            int[] cnt = new int[m + 1];

            for (int h : heights) {
          
   
                ++cnt[h];
            }

            int idx = 0, ans = 0;

            for (int i = 1; i <= m; i++) {
          
   
                for (int j = 1; j <= cnt[i]; j++) {
          
   
                    if (heights[idx] != i) {
          
   
                        ans++;
                    }
                    idx++;
                }
            }

            return ans;

        }
    }
}

复杂度分析

时间复杂度：O(n + C)，其中 n 是数组 heights 的长度，C是数组heights 中的最大值。即为计数排序需要的时间。 空间复杂度：O©，即为计数排序需要的空间。

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