Leetcode 870. Advantage Shuffle

Tyan

发布于 2019-05-25 22:49:14

3130

发布于 2019-05-25 22:49:14

文章被收录于专栏：SnailTyanSnailTyan

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;
        }
    }
};