Leetcode 870. Advantage Shuffle | | Leetcode 870. Advantage Shuffle 文章作者:Tyan博客:noahsnail.com | CSDN | 简书 1. Description 2. Solution Version 1 12345678910111213141516171819202122232425class Solution {public: vector<int> advantageCount(vector<int>& A, vector<int>& B) { vector<int> result(A.size()); sort(A.begin(), A.end()); for(int i = 0; i < B.size(); i++) { int index = findLargerNumber(A, B[i]); result[i] = A[index]; A.erase(A.begin() + index); } return result; }private: int findLargerNumber(vector<int>& A, int target) { if(A[A.size() - 1] <= target) { return 0; } for(int i = 0; i < A.size(); i++) { if(A[i] > target) { return i; } } }}; Version 2 1234567891011121314151617181920212223242526272829303132333435class Solution {public: vector<int> advantageCount(vector<int>& A, vector<int>& B) { vector<int> result(A.size()); sort(A.begin(), A.end()); for(int i = 0; i < B.size(); i++) { int index = findLargerNumber(A, B[i]); result[i] = A[index]; A.erase(A.begin() + index); } return result; }private:private: int findLargerNumber(vector<int>& A, int target) { int left = 0; int right = A.size() - 1; while(left < right) { int middle = (left + right) / 2; if(A[middle] <= target) { left = middle + 1; } else { right = middle; } } if(A[right] <= target) { return 0; } else { return right; } }}; Reference https://leetcode.com/problems/advantage-shuffle/description/ 如果有收获,可以请我喝杯咖啡! 赏 微信打赏 支付宝打赏