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| 1 | +package com.bruis.algorithminjava.algorithm.leetcode; |
| 2 | + |
| 3 | +import java.util.Arrays; |
| 4 | + |
| 5 | +/** |
| 6 | + * 逆序对 |
| 7 | + * <p> |
| 8 | + * url: https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/ |
| 9 | + * |
| 10 | + * @author LuoHaiYang |
| 11 | + */ |
| 12 | +public class ReversePairs { |
| 13 | + |
| 14 | + /* ================================ 解法一 ================================*/ |
| 15 | + |
| 16 | + /** |
| 17 | + * 暴力解法O(n^2),超时 |
| 18 | + * |
| 19 | + * @param nums |
| 20 | + * @return |
| 21 | + */ |
| 22 | + public int reversePairs2(int[] nums) { |
| 23 | + int n = nums.length; |
| 24 | + if (n < 2) { |
| 25 | + return 0; |
| 26 | + } |
| 27 | + int reverseNum = 0; |
| 28 | + for (int i = 0; i < n; i++) { |
| 29 | + for (int j = i + 1; j < n; j++) { |
| 30 | + if (nums[i] > nums[j]) { |
| 31 | + reverseNum++; |
| 32 | + } |
| 33 | + } |
| 34 | + } |
| 35 | + return reverseNum; |
| 36 | + } |
| 37 | + |
| 38 | + /* ================================ 解法二 ================================*/ |
| 39 | + |
| 40 | + /** |
| 41 | + * 使用自顶向下的归并排序算法计算逆序对,用来额外的空间。 |
| 42 | + * |
| 43 | + * @param nums |
| 44 | + * @return |
| 45 | + */ |
| 46 | + public int reversePairs(int[] nums) { |
| 47 | + int n = nums.length; |
| 48 | + if (n < 2) { |
| 49 | + return 0; |
| 50 | + } |
| 51 | + return getReversePairs(nums); |
| 52 | + } |
| 53 | + |
| 54 | + private int getReversePairs(int[] nums) { |
| 55 | + int n = nums.length; |
| 56 | + return getReversePairs(nums, 0, n - 1); |
| 57 | + } |
| 58 | + |
| 59 | + private int getReversePairs(int[] nums, int left, int right) { |
| 60 | + if (left >= right) { |
| 61 | + return 0; |
| 62 | + } |
| 63 | + int result = 0; |
| 64 | + int mid = (left + right) / 2; |
| 65 | + result += getReversePairs(nums, left, mid) + getReversePairs(nums, mid + 1, right) + reversePairs(nums, left, mid, right); |
| 66 | + return result; |
| 67 | + } |
| 68 | + |
| 69 | + private int reversePairs(int[] nums, int left, int mid, int right) { |
| 70 | + int[] aux = Arrays.copyOfRange(nums, left, right + 1); |
| 71 | + int i = left, j = mid + 1; |
| 72 | + |
| 73 | + int res = 0; |
| 74 | + |
| 75 | + for (int k = left; k <= right; k++) { |
| 76 | + if (i > mid) { |
| 77 | + nums[k] = aux[j - left]; |
| 78 | + j++; |
| 79 | + } else if (j > right) { |
| 80 | + nums[k] = aux[i - left]; |
| 81 | + i++; |
| 82 | + } else if (aux[i - left] <= aux[j - left]) { |
| 83 | + nums[k] = aux[i - left]; |
| 84 | + i++; |
| 85 | + } else { |
| 86 | + nums[k] = aux[j - left]; |
| 87 | + j++; |
| 88 | + res += (mid - i) + 1; |
| 89 | + } |
| 90 | + } |
| 91 | + return res; |
| 92 | + } |
| 93 | + |
| 94 | + /* ================================ 题解三(优化) ================================*/ |
| 95 | + |
| 96 | + /** |
| 97 | + * @param nums |
| 98 | + * @return |
| 99 | + */ |
| 100 | + public int reversePairs3(int[] nums) { |
| 101 | + if (nums == null || nums.length < 2) |
| 102 | + return 0; |
| 103 | + int[] temp = new int[nums.length]; |
| 104 | + System.arraycopy(nums, 0, temp, 0, nums.length); |
| 105 | + |
| 106 | + int count = mergeCount(nums, temp, 0, nums.length - 1); |
| 107 | + return count; |
| 108 | + } |
| 109 | + |
| 110 | + private int mergeCount(int[] nums, int[] temp, int start, int end) { |
| 111 | + if (start >= end) { |
| 112 | + return 0; |
| 113 | + } |
| 114 | + |
| 115 | + int mid = (start + end) >> 1; |
| 116 | + int left = mergeCount(temp, nums, start, mid); |
| 117 | + int right = mergeCount(temp, nums, mid + 1, end); |
| 118 | + int count = 0; |
| 119 | + |
| 120 | + //merge() |
| 121 | + int i = mid;//遍历左区域指针 |
| 122 | + int j = end;//遍历右区域指针 |
| 123 | + |
| 124 | + int k = end;//临时区域指针 |
| 125 | + while (i >= start && j >= mid + 1) { |
| 126 | + if (nums[i] > nums[j]) { |
| 127 | + count += j - mid; |
| 128 | + temp[k--] = nums[i--]; |
| 129 | + } else { |
| 130 | + temp[k--] = nums[j--]; |
| 131 | + } |
| 132 | + } |
| 133 | + //如果还有剩下没遍历的 |
| 134 | + while (i >= start) |
| 135 | + temp[k--] = nums[i--]; |
| 136 | + while (j >= mid + 1) |
| 137 | + temp[k--] = nums[j--]; |
| 138 | + |
| 139 | + return count + left + right; |
| 140 | + } |
| 141 | + |
| 142 | + public static void main(String[] args) { |
| 143 | + ReversePairs reversePairs = new ReversePairs(); |
| 144 | + int[] nums = {7, 5, 6, 4}; |
| 145 | + //int[] nums = {1,3,2,3,1}; |
| 146 | + System.out.println(reversePairs.reversePairs3(nums)); |
| 147 | + } |
| 148 | + |
| 149 | + |
| 150 | +} |
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