Leetcode-1051.Height Checker
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:
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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
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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
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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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