分治-归并系列一＞翻转对

题目：

链接: link

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这题和逆序对区别点就是，要找到前一个元素是后一个元素的2倍
先找到目标值再，继续堆排序

解析：

策略一：

代码：

class Solution {
    int[] tmp;
    public int reversePairs(int[] nums) {
        int n = nums.length;
        tmp = new int[n];
        return mergesort(nums,0,n-1);
    }
    
    private int mergesort(int[] nums, int left, int right){
        int ret = 0;
        if(left >= right) return 0;

        int mid = (right + left) / 2;
        //左右两边找翻转对
        ret += mergesort(nums,left,mid);
        ret += mergesort(nums,mid+1,right);

        //一左一右找翻转对: 降序版本
        //输入数组中的所有数字都在32位整数的表示范围内:改为：2.0*nums[cur2]
        int cur1 = left, cur2 = mid+1, i = 0;
        while(cur1 <= mid && cur2 <= right){
            if(nums[cur1] <= 2.0*nums[cur2]){
                cur2++;
            }else {
                ret += right - cur2 + 1;
                cur1++;
            }
            if(cur2 > right) break;
        }

        //排序：
        cur1 = left; cur2 = mid+1;
        while(cur1 <= mid && cur2 <= right) 
            tmp[i++] = nums[cur1] <= nums[cur2]? nums[cur2++] : nums[cur1++];

        while(cur1 <= mid) tmp[i++] = nums[cur1++];
        while(cur2 <= right) tmp[i++] = nums[cur2++];
        
        //放回原数组：
        for(int j = left; j <= right; j++){
            nums[j] = tmp[j-left];
        }

        return ret;
    }
}

策略二：

代码：

class Solution {
    int[] tmp;
    public int reversePairs(int[] nums) {
        int n = nums.length;
        tmp = new int[n];
        return mergesort(nums,0,n-1);
    }

    一左一右找翻转对: 升序版本:
     private int mergesort(int[] nums, int left, int right){
        int ret = 0;
        if(left >= right) return 0;

        int mid = (right + left) / 2;
        //左右两边找翻转对
        ret += mergesort(nums,left,mid);
        ret += mergesort(nums,mid+1,right);

        //一左一右找翻转对: 升序版本
        //输入数组中的所有数字都在32位整数的表示范围内:改为：2.0*nums[cur2]
        int cur1 = left, cur2 = mid+1, i = 0;
        while(cur1 <= mid && cur2 <= right){
            if(nums[cur1] / 2.0 <= nums[cur2]){
                cur1++;
            }else {
                ret += mid - cur1 + 1;
                cur2++;
            }
            if(cur1 > mid) break;
        }

        //排序：
        cur1 = left; cur2 = mid+1;
        while(cur1 <= mid && cur2 <= right) 
            tmp[i++] = nums[cur1] <= nums[cur2]? nums[cur1++] : nums[cur2++];

        while(cur1 <= mid) tmp[i++] = nums[cur1++];
        while(cur2 <= right) tmp[i++] = nums[cur2++];
        
        //放回原数组：
        for(int j = left; j <= right; j++){
            nums[j] = tmp[j-left];
        }

        return ret;
    }
}