Description
Given an array nums, we call (i, j) an important reverse pair if i < j and nums[i] > 2*nums[j].
You need to return the number of important reverse pairs in the given array.
Example
Example1:
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| Input: [1,3,2,3,1]
Output: 2
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Example2:
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| Input: [2,4,3,5,1]
Output: 3
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Solution
- Time Complexity:
- Space Complexity:
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| class Solution {
public int reversePairs(int[] nums) {
if (nums == null || nums.length == 0) return 0;
return mergeSort(nums, 0, nums.length - 1);
}
private int mergeSort(int[] nums, int left, int right){
if (left >= right) return 0;
int mid = left + (right - left) / 2;
int res = 0;
res += mergeSort(nums, left, mid);
res += mergeSort(nums, mid + 1, right);
res += merge(nums, left, mid, right);
return res;
}
private int merge(int[] nums, int left, int mid, int right){
int count = 0;
int p = left, q = mid + 1;
while (p <= mid && q <= right){
if ((long) nums[p] > 2 * (long) nums[q]){
count += mid - p + 1;
q++;
}
else{
p++;
}
}
int[] temp = new int[right - left + 1];
p = left;
q = mid + 1;
int index = 0;
while (p <= mid && q <= right){
if (nums[p] > nums[q]){
temp[index++] = nums[q++];
}
else{
temp[index++] = nums[p++];
}
}
while (p <= mid){
temp[index++] = nums[p++];
}
while (q <= right){
temp[index++] = nums[q++];
}
System.arraycopy(temp, 0, nums, left, right - left + 1);
return count;
}
}
|