剑指 Offer 51. 数组中的逆序对
题目链接
https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/
代码连接:
题目解析:
- 方法1:归并排序
- 方法2:离散化树状数组
官方题解:https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/solution/shu-zu-zhong-de-ni-xu-dui-by-leetcode-solution/
package leetcode.offer100.arrays;
import base.Utils;
/**
* 题目:剑指 Offer 51. 数组中的逆序对
* 题目连接:https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/
*
* 方法1:归并排序
* 方法2:离散化树状数组
* 官方题解:https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/solution/shu-zu-zhong-de-ni-xu-dui-by-leetcode-solution/
*
*/
public class HardReversePairs {
/**
* 1. 使用归并排序解题
* @param nums
* @return
*/
public static int reversePairs(int[] nums) {
int len = nums.length;
if (len < 2) {
return 0;
}
int[] temp = new int[len];
return countReversePairs(nums, temp, 0, len - 1);
}
private static int countReversePairs(int[] nums, int[] temp, int left, int right) {
if (left == right) {
return 0;
}
int mid = Utils.midNumber(left, right);
int leftPairs = countReversePairs(nums, temp, left, mid);
int rightPairs = countReversePairs(nums, temp, mid + 1, right);
if (nums[mid] <= nums[mid + 1]) {
return leftPairs + rightPairs;
}
int crossPairs = countAndMerge(nums, temp, left, mid, right);
return leftPairs + crossPairs + rightPairs;
}
private static int countAndMerge(int[] nums, int[] temp, int left, int mid, int right) {
for (int i = left; i <= right ; i++) {
temp[i] = nums[i];
}
int i = left, j = mid + 1;
int count = 0;
for (int k = left; k <= right; k++) {
if (i == mid + 1) {
nums[k] = temp[j++];
} else if (j == right + 1) {
nums[k] = temp[i++];
} else if (temp[i] <= temp[j]) {
nums[k] = temp[i++];
} else {
nums[k] = temp[j++];
count += (mid - i + 1);
}
}
return count;
}
public static void main(String[] args) {
}
}