Best Time to Buy and Sell Stock 最佳时间买入卖出股票（一次买入卖出） @LeetCode

题目：

最佳时间买入卖出股票：你有一个数组保存了股票在第i天的价钱，现在你只能进行一次买入卖出，如何赚的最多

思路：

min记录最小买入价

maxProfit记录最大利润

遍历array，不断更新最小买入价，计算更新最大利润

package Level2;/** * Best Time to Buy and Sell Stock  *  * Say you have an array for which the ith * element is the price of a given stock on day i. *  * If you were only permitted to complete at most one transaction (ie, buy one * and sell one share of the stock), design an algorithm to find the maximum * profit. */public class S121 {public static void main(String[] args) {}public int maxProfit(int[] prices) {// 当没有任何prices信息的情况if(prices.length == 0){return 0;}        int min = prices[0];// 记录最小买入价的index        int maxProfit = 0;// 记录最大profit                for(int i=1; i<prices.length; i++){        // 更新最小买入价        if(prices[i] < min){        min = prices[i];        }                // 计算当前利润        int currentProfit = prices[i] - min;        // 如果当前利润超过最大利润，更新最大利润        if(currentProfit > maxProfit){        maxProfit = currentProfit;        }        }                return maxProfit;    }}

public class Solution {    public int maxProfit(int[] prices) {        if(prices.length == 0){            return 0;        }        int minBuy = prices[0];        int maxProfit = 0;                for(int i=0; i<prices.length; i++){            minBuy = Math.min(minBuy, prices[i]);            maxProfit = Math.max(maxProfit, prices[i]-minBuy);        }        return maxProfit;    }}

采用“局部最优和全局最优解法”：

思路是维护两个变量，一个是到目前为止最好的交易，另一个是在当前一天卖出的最佳交易（也就是局部最优）。递推式是local[i+1]=max(local[i]+prices[i+1]-price[i],0), global[i+1]=max(local[i+1],global[i])。这样一次扫描就可以得到结果，时间复杂度是O(n)。而空间只需要两个变量，即O(1)。

http://blog.csdn.net/linhuanmars/article/details/23162793

有两点值得注意：

1 在第i天买入，在i+1天卖出，再在i+1天买入，在i+2天卖出，等效于在i天买入在i+2天卖出：p[i+2]-p[i] = -p[i]+p[i+1]-p[i+1]+p[i+2]

2 如果计算出没两天的差值：diff1, diff2, diff3 则此题就转化为maximum subarray的问题

public class Solution {    public int maxProfit(int[] prices) {        int len = prices.length;        if(len == 0){            return 0;        }        int[] local = new int[len+1];        int[] global = new int[len+1];                local[0] = 0;        global[0] = 0;        for(int i=1; i<len; i++) {            int diff = prices[i] - prices[i-1];            local[i] = Math.max(local[i-1]+diff, Math.max(diff, 0));            global[i] = Math.max(global[i-1], local[i]);        }                return global[len-1];    }}