493 Reverse Pairs
Problem:
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.
Example1:
Input: [1,3,2,3,1]
Output: 2Example2:
Input: [2,4,3,5,1]
Output: 3Note: The length of the given array will not exceed 50,000. All the numbers in the input array are in the range of 32-bit integer.
Solutions:
public class Solution {
public int reversePairs(int[] nums) {
HashMap<Integer, Integer> index = new HashMap<Integer, Integer>();
int[] sorted = new int[nums.length];
int[] bit = new int[nums.length + 1];
for (int i = 0; i < nums.length; i ++) {
sorted[i] = nums[i];
}
Arrays.sort(sorted);
for (int i = 0; i < sorted.length; i ++) {
index.put(sorted[i], i + 1);
}
int res = 0;
for (int i = nums.length - 1; i >= 0; i --) {
int j = index(sorted, (double) nums[i]/2.0);
int tmp = getSum(bit, j);
res += tmp;
update(bit, index.get(nums[i]));
}
return res;
}
private int index(int[] sorted, double val) {
int left = 0, right = sorted.length - 1;
while (left <= right) {
int mid = (right - left) / 2 + left;
if (sorted[mid] >= val) {
right = mid - 1;
}
else {
left = mid + 1;
}
}
return right + 1;
}
private int getSum(int[] bit, int i) {
int res = 0;
while (i > 0) {
res += bit[i];
i -= (i&(-i));
}
return res;
}
private void update(int[] bit, int i) {
while (i < bit.length) {
bit[i] ++;
i += (i&(-i));
}
}
}Last updated