Leetcode 870. Advantage Shuffle

文章作者:Tyan
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1. Description

Advantage Shuffle

2. Solution

  • Version 1
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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
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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/
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