LeetCode: Best Time to Buy and Sell Stock III

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

class Solution {
public:
int maxProfit(vector<int> &prices) {
if(prices.size() == 0) return 0;
int maxProfit = 0;
int maxp1 = 0;
int maxp2 = 0;
int minp1 = prices[0];
int minp2 = INT_MAX;
for(int i = 1;i < prices.size();i++){
int diff1 = prices[i] - minp1;
if(diff1 < 0){
minp1 = prices[i];
}else if(diff1 > maxp1){
maxp1 = diff1;
}

if(i < prices.size() - 1){
for(int j = i; j < prices.size();j++){
int diff2 = prices[j] - minp2;
if(diff2 < 0){
minp2 = prices[j];
}else if(diff2 > maxp2){
maxp2 = diff2;
}
}
}else{
maxp2 = 0;
}
maxProfit = (maxp1 + maxp2) > maxProfit ? maxp1 + maxp2:maxProfit;
}
return maxProfit;
}
};

Submission Result: Time Limit Exceeded

在大测试数据面前，还是过不了。

another solution googled can be accepted by the LeetCode

public class Solution {
public int maxProfit(int[] prices) {
// Start typing your Java solution below
// DO NOT write main() function
int min = Integer.MAX_VALUE, max=Integer.MIN_VALUE;
int [] forward = new int[prices.length],
backward = new int[prices.length];

for(int i=0;i<prices.length;i++){
if(prices[i]>min){
forward[i]=Math.max(prices[i]-min,forward[i-1]);
}else{
if(i>0) forward[i]=forward[i-1];
}
min=Math.min(prices[i],min);

if(prices[prices.length-1-i]<max){
backward[prices.length-1-i]=Math.max(max-prices[prices.length-1-i],backward[prices.length-i]);
}else{
if(i>0) backward[prices.length-1-i]=backward[prices.length-i];
}
max=Math.max(prices[prices.length-1-i],max);
}

int res = 0;

for(int i=0;i<prices.length;i++){
res=Math.max(forward[i]+backward[i],res);
}
return res;
}
}