leetcode-Best Time to Buy and Sell Stock(2014.2.16)
2014-04-19 23:06
316 查看
用两层遍历的方法超时:
class Solution {
public:
int maxProfit(vector<int> &prices) {
int maxPro=0;
for(int i=0;i<prices.size();i++){
for(int j=i+1;j<prices.size();j++){
int m=prices[j]-prices[i];
if(m>maxPro) maxPro=m;
}
}
return maxPro;
}
};
改用记录最低值得方法,一次遍历则完成,免去大量重复计算,注意判断长度以确保程序的健壮性:
class Solution {
public:
int maxProfit(vector<int> &prices) {
if(prices.size()<2) return 0;
int maxPro=0;
int low=prices[0];
for(int i=0;i<prices.size();i++){
int m=prices[i]-low;
if(prices[i]<low) low=prices[i];
if(m>maxPro) maxPro=m;
}
return maxPro;
}
};
class Solution {
public:
int maxProfit(vector<int> &prices) {
int maxPro=0;
for(int i=0;i<prices.size();i++){
for(int j=i+1;j<prices.size();j++){
int m=prices[j]-prices[i];
if(m>maxPro) maxPro=m;
}
}
return maxPro;
}
};
改用记录最低值得方法,一次遍历则完成,免去大量重复计算,注意判断长度以确保程序的健壮性:
class Solution {
public:
int maxProfit(vector<int> &prices) {
if(prices.size()<2) return 0;
int maxPro=0;
int low=prices[0];
for(int i=0;i<prices.size();i++){
int m=prices[i]-low;
if(prices[i]<low) low=prices[i];
if(m>maxPro) maxPro=m;
}
return maxPro;
}
};
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