Sword refers to Offer 51. Reversed pairs in the array (Merge)

LeetCode: Sword refers to Offer 51. Reversed pairs in the array

Merge sort

Old Rules Trilogy

1. Minute
2. Logical processing
3. And (sort)

When the logic is processed for the first time, it has already been sorted.

AC Code

class Solution {
    
    
    public int reversePairs(int[] nums) {
    
    
        // 
        int len = nums.length;
        if(len < 2) return 0;
        int ans = devide(nums, 0, len - 1);
        return ans;
    }

    public int devide(int[] nums, int left, int right){
    
    
        if(left >= right) return 0;
        int mid = left + (right - left) / 2;
        // 分 
        int a = devide(nums, left, mid);
        int b = devide(nums, mid + 1, right);
        int res = a + b;

        // 处理
        // 前面一个数字大于后面的数字  >>  结果 + 1
        int s1 = left, s2 = mid + 1;
        while(s1 < mid + 1) {
    
    
            // 最小的两个数，已经排好序了.
            while(s2 <= right && nums[s1] > nums[s2]) {
    
    
                s2++;
            }

            res += s2 - (mid + 1);
            // 下一个数
            s1++;
        }


        // 并  >> 排序
        int[] arr = new int[right - left + 1];
        int l = left, r = mid + 1, idx = 0;
        while(l < mid + 1 || r <= right) {
    
    
            if(l >= mid + 1) {
    
    
                arr[idx++] = nums[r++];
            } else if(r > right) {
    
    
                arr[idx++] = nums[l++];
            } else {
    
    
                if (nums[l] < nums[r]) arr[idx++] = nums[l++];
                else arr[idx++] = nums[r++];
            }
        }

        // 覆盖原数组
        for(int i = 0; i < idx; i++) {
    
    
            nums[left + i] = arr[i];
        }

        return res;
    }

}