Leetcode 870. Advantage Shuffle

时间:2022-06-22
本文章向大家介绍Leetcode 870. Advantage Shuffle,主要内容包括其使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。

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文章作者:Tyan 博客:noahsnail.com | CSDN | 简书

1. Description

2. Solution

  • Version 1
class 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
class 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

  1. https://leetcode.com/problems/advantage-shuffle/description/