对于数组A[0,1,2,3,4,...,k]，求得0<=i < j < k，且使得A[j] - A[i]为最大值。
#include <iostream>
#include <vector>
#include "math.h"
using namespace std;
class Solution{
public:
bool maxSum(vector<int> & array, int & result){
if(array.size() <= 1)
return false;
else if(array.size() == 2){
result = array[1] - array[0];
return true;
}
else{
int preMin = array[0], tempMax = 0;
for(int i = 1; i < array.size(); i ++){
if(array[i] < preMin){
preMin = array[i];
}
else
tempMax = max(tempMax, array[i] - preMin);
}
result = tempMax;
return true;
}
}
};
int main() {
cout << "test\n";
Solution sol;
int arrayTemp[] = {2, 3, 8, 1, 7, 9};
vector<int> array(arrayTemp, arrayTemp + 6);
int result;
sol.maxSum(array,result);
cout << result << endl;
return 0;
}
