315. Count of Smaller Numbers After Self

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].

 

Solution1: tranverse the nums array starting from the last number, insert the number into the already ordered vector t, the index of the number in t is just counts[i]. Here use the binary search to implement the insertion.

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

Solution 2: line 6 use the STL function lower_bound to find the positon of the insertion, however, the time complexity is O(n^2) because of the insertion.

class Solution {
public:
    vector<int> countSmaller(vector<int>& nums) {
        vector<int> t, res(nums.size());
        for (int i = nums.size() - 1; i >= 0; i--) {            
            res[i] = lower_bound(t.begin(),t.end(),nums[i])-t.begin();
            t.insert(t.begin() + res[i], nums[i]);
        }
        return res;
    }
};

Solution 3:merge sort  http://blog.csdn.net/qq508618087/article/details/51320926