LeetCode 315. 计算右侧小于当前元素的个数

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example 1:

Input: nums = [5,2,6,1]
Output: [2,1,1,0]
Explanation:
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.

Constraints:

• 0 <= nums.length <= 10^5
• -10^4 <= nums[i] <= 10^4

解题思路

1. 使用二分查找优化的插入排序
2. 二叉搜索树
3. 归并排序
4. 树状数组
5. 线段树

参考解析

参考代码

/*
 * @lc app=leetcode id=315 lang=cpp
 *
 * [315] Count of Smaller Numbers After Self
 */

// @lc code=start
class Solution {
public:
    // Insertion Sort with binary search optimization
    vector<int> countSmaller(vector<int>& nums) {
        if (nums.empty()) return {};
        size_t n = nums.size();
        vector<int> res(n);
        res[n-1] = 0;
        for (int i=n-2; i>=0; i--) {
            // find the 1st elements less than val
            int val = nums[i];
            int l = i+1, r = n;
            while (l < r) {
                int m = l + (r - l) / 2;
                if (nums[m] >= val) {
                    l = m + 1;
                } else {
                    r = m;
                }
            } // upper_bound in descending array
            // return l;
            for (int j = i; j + 1 < l; j++) {
                nums[j] = nums[j+1];
            }
            nums[l-1] = val;
            res[i] = n-l;
        }
        return res;
    } // AC, runtime beats 5.07 % of cpp submissions
};
// @lc code=end

注意：使用开区间，这样区间只有两个元素的时候选择的mid是第二个元素，避免了nums[m] < val时更新r = m的死循环。

upper_bound 和 lower_bound 的实现参考：

1. GeeksforGeeks
2. StackOverflow

标签：upper,lower,smaller,val,nums,int,bound,right
来源： https://www.cnblogs.com/zhcpku/p/14318805.html