123:Best Time to Buy and Sell Stock III【数组】【DP】

/*题意：一个数组，第i元素表示第i天股票的价格，允许最多买卖两次，求最大利润 */

/**
*思路：用currProfit数组记录截止当日的最大利润，从头到尾扫描一遍数组即可获得
*      currProfit = max(currPorfit[i], prices[i]-low)
*
*      用futureProfit数组记录当日以后的最大利润，从尾到头扫描可得
*      futureProfit = max(futureProfit[i], high-prices[i])
*
*      计算总利润：anx = max(ans, currProfit[i]+futureProfit[i])
*/

class Solution {
public:
int maxProfit(vector<int> &prices) {
int len = prices.size();
if(len <= 1) return 0;
int *currProfit = new int[len];
int *futureProfit = new int[len];
currProfit[0] = futureProfit[len-1] = 0;

//记录截止当日的最大利润
int low = prices[0];
for(int i = 1; i < len; i ++) {
low = min(low, prices[i]);
currProfit[i] = max(currProfit[i-1], prices[i]-low);
}

//计算该日以后的最大利润
int high = prices[len-1];
for(int i = len-2; i >= 0; i --) {
high = max(high, prices[i]);
futureProfit[i] = max(futureProfit[i+1], high-prices[i]);
}

//合计
int ans = 0;
for(int i = 0; i < len; i ++)
ans = max(ans, currProfit[i] + futureProfit[i]);
return ans;
}
};