LeetCode 315.Count of Smaller Numbers After Self

题目

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:

Given nums = [5, 2, 6, 1]
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.

Return the array [2, 1, 1, 0].

难易度：Hard

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思路

这道题是要求数出数组中每个数字后面比它小的数有几个，然后返回一个新的数组存放这些个数。最直接的方法就是一个一个的遍历，这样复杂度要O(n^2)。所以用另外一个思路，利用二分排序，将数组从最后一个插入到一个新的数组中，这样每个数字插入的位置的序号就是后面比它小的元素的个数。

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代码

class Solution {public:    vector<int> countSmaller(vector<int>& nums) {        int n = nums.size();        vector<int> arr,count(n);        for(int i = n-1; i >= 0; --i )        {            int left = 0;            int right = arr.size();            while(left < right){                int mid = left+(right-left)/2;                if(nums[i] <= arr[mid])                {                    right = mid;                }                else                 {                    left = mid+1;                }            }            count[i] = right;            arr.insert(arr.begin()+right, nums[i]);        }        return count;    }};

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注：
1. vector.insert(vector.begin()+i,a); 在vector第i个元素后插入a
2. 在进行二分排序时，注意边界值的判断和选择