Time to Buy and Sell Stock系列

Best Time to Buy and Sell Stock I

给一个数prices[]，prices[i]代表股票在第i天的售价，求出只做一次交易(一次买入和卖出)能得到的最大收益。
只需要找出最大的差值即可，即 max(prices[j] – prices[i]) ，i < j。一次遍历即可，在遍历的时间用遍历low记录 prices[o….i] 中的最小值，就是当前为止的最低售价，时间复杂度为 O(n)。

public class Solution {    public int maxProfit(int[] prices) {        if(prices.length == 0) return 0;        int low = prices[0];        int ans = 0;        for(int i=1; i<prices.length; i++){            if(prices[i] < low) low = prices[i];            else if(prices[i] - low > ans) ans = prices[i] - low;        }        return ans;    }}

Best Time to Buy and Sell Stock II

贪心算法：
此题和上面一题的不同之处在于不限制交易次数。也是一次遍历即可，只要可以赚就做交易。

public class Solution {    public int maxProfit(int[] prices) {      if(prices.length == 0) return 0;      int ans = 0;      for(int i=1; i<prices.length; i++){          if(prices[i] > prices[i-1])            ans += prices[i]-prices[i-1];      }      return ans;    }}

Best Time to Buy and Sell Stock III

此题是限制在两次交易内，相对要难一些。容易想到的解决办法是，把prices[] 分成两部分prices[0…m] 和 prices[m…length] ，分别计算在这两部分内做交易的做大收益。

class Solution {public:    /**     * 计算你能获得的最大收益     *      * @param prices Prices[i]即第i天的股价     * @return 整型     */    int calculateMax(vector<int> prices) {        vector<int> pre(prices.size(),0);        vector<int> post(prices.size(), 0);        int profix = 0;        int low = prices[0];        for (int i = 0; i < prices.size(); i++)        {            if (prices[i] < low)                 low = prices[i];            if (profix < prices[i] - low)                 profix = prices[i] - low;            pre[i] = profix;        }        int hight = prices[prices.size() - 1];        profix = 0;        for (int i = prices.size() - 1; i >= 0; i--)        {            if (prices[i] > hight)                 hight = prices[i];            if (profix < hight - prices[i])                profix = hight - prices[i];            post[i] = profix;        }        profix = 0;        for (int i = 0; i < prices.size(); i++)        {            if (profix < pre[i] + post[i])                profix = pre[i] + post[i];        }        return profix;    }};