493. Reverse Pairs

Hard

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: 2

Example2:

Input: [2,4,3,5,1]
Output: 3

Note:

1. The length of the given array will not exceed 50,000.
2. All the numbers in the input array are in the range of 32-bit integer.

思路：这道题和求数组最小和思路一样，采用归并排序，归并排序比普通方法快的本质是不会浪费之前的组内比较，只需跨组比较。

class Solution {
    public int reversePairs(int[] nums) {
        if(nums==null || nums.length<2) return 0;
        return mergeSort(nums,0,nums.length-1);
    }
    public int mergeSort(int[] nums,int start,int end){
        if(start==end) return 0;
        int mid = start + ((end-start)>>1);
        int count = mergeSort(nums,start,mid) + mergeSort(nums,mid+1,end);
        //核心逻辑部分，其余部分是归并排序模板
        for(int i=start,j=mid+1;i<mid+1;i++){
            while(j<=end&&nums[i]/2.0>nums[j]) j++;//nums[i]/2.0防止溢出
            count += j - (mid + 1);
        }

        merge(nums,start,mid,end);
        return count;
    }
    public void merge(int[] nums,int start,int mid,int end){
        int[] helper = new int[end-start+1];
        int i = 0;
        int p1 = start;
        int p2 = mid + 1;
        while(p1<=mid&&p2<=end){
            helper[i++] = nums[p1]<nums[p2]?nums[p1++]:nums[p2++];
        }
        while(p1<=mid){
            helper[i++] = nums[p1++];
        }
        while(p2<=end){
            helper[i++] = nums[p2++];
        }
        for(int j=0;j<helper.length;j++){
            nums[start+j] = helper[j];
        }
    }
}

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