315. Count of Smaller Numbers After Self Hard
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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]
.
思路:由于题目测试量很大,普通遍历查找结果会超时,所以采用二分查找,需要使用一个辅助数组,从原数组的最后一位开始,对辅助数组进行二分查找,得到此数在辅助数组中以小到大所在的位置的下标,这个下标就是题目所求小于此数的个数,之后按照从小到大的顺序将此数插入到辅助数组中,再次进行新的查找。最后得到的一组下标值就是所需结果。
int binary(vector<int>temp, int num) { int begin = 0, end = temp.size() - 1; while (begin <= end) { int mid = (begin + end) / 2; if (temp[mid] >= num) { end = mid - 1; } else { begin = mid + 1; } } return begin;}class Solution {public: vector<int> countSmaller(vector<int>& nums) { vector<int>temp; vector<int>result; for (int i = nums.size() - 1; i >= 0; i--) { int set = binary(temp, nums[i]); temp.insert(temp.begin() + set, nums[i]); result.insert(result.begin(), set); } return result; }};
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