在数组中的两个数字,如果前面一个数字大于后面的数字,则这两个数字组成一个逆序对。输入一个数组,求出这个数组中的逆序对的总数。
示例 1:
输入: [7,5,6,4]
输出: 5
限制:
0 <= 数组长度 <= 50000
归并排序
class Solution {
public int reversePairs(int[] nums) {
int len = nums.length;
int[] copy = new int[len];
int[] temp = new int[len];
if(len < 2) return 0;
for(int i = 0; i < len; i++){
copy[i] = nums[i];
}
return mergeSort(copy, 0, len - 1, temp);
}
public int mergeSort(int[] nums, int left, int right, int[] temp){
if(left == right) return 0;
int mid = left + (right - left) / 2;
int leftPairs = mergeSort(nums, left, mid, temp);
int rightPairs = mergeSort(nums, mid + 1, right, temp);
if(nums[mid] <= nums[mid + 1]){
return leftPairs + rightPairs;
}
int crossPairs = mergeCount(nums, left, mid, right, temp);
return leftPairs + rightPairs + crossPairs;
}
public int mergeCount(int[] nums, int left, int mid, int right, int[] temp){
for(int i = left; i <= right; i++){
temp[i] = nums[i];
}
int i = left;
int j = mid + 1;
int count = 0;
for(int k = left; k <= right; k++){
if(i == mid + 1){
nums[k] = temp[j];
j++;
}else if(j == right + 1){
nums[k] = temp[i];
i++;
}else if(temp[i] <= temp[j]){
nums[k] = temp[i];
i++;
}else{
nums[k] = temp[j];
j++;
count += (mid - i + 1);
}
}
return count;
}
}
树状数组
待补