leetcode-121-Best Time to Buy and Sell Stock
2017-02-14 13:57
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问题
题目:[leetcode-121]思路
蛮力法,计算所有可能的情形。O(N2)的复杂度。代码(TLE)
class Solutiocn { public: int maxProfit(vector<int>& prices) { int sz = prices.size(); if(!sz) return 0; int ans = 0; for(int i = 0; i < sz; ++i){ for(int j = i + 1; j < sz; ++j){ int profix = prices[j] - prices[i]; ans = std::max(ans, profix); } } return ans; } };
思路1
看了最长连续字段和的代码。有点启发。首先这两个问题本质都是找一个区间,虽然后面这个其实不是找区间。但是需要找到start,end两个下标。既然,前面的题目都可以解。这个题我觉得应该也可以。
对比之前的思路,一定要考虑当前数值和之前数值的关系,我就是按着这个方向才想到了解决办法。O(N)的复杂度。DP求解。
当然,我下面你的做法还是标记了买入点和卖出点。最大字段和是没有明确的标记,它是当dp[i-1]<=0时,重新计算。就相当于标记了开始点。
定义:dp[i]表示在第i点卖出时的利润,则状态转移函数如下:
dp[i]={prices[i]−prices[buy]−1 and buy=i, prices[buy]<prices[i], prices[buy]≥prices[i] (1)
代码1
class Solution { public: int maxProfit(vector<int>& prices) { int sz = prices.size(); if(!sz) return 0; int buy = 0; int max = 0; std::vector<int> dp(sz, int()); dp[0] = -1; for(int i = 1; i < sz; ++i){ if(prices[buy] < prices[i]) dp[i] = prices[i] - prices[buy]; else{ buy = i; dp[i] = -1; } max = std::max(max, dp[i]); } return max; } };
代码2
省去空间的开销。class Solution { public: int maxProfit(vector<int>& prices) { int sz = prices.size(); if(!sz) return 0; int buy = 0; int max = 0; int profit = -1; for(int i = 1; i < sz; ++i){ if(prices[buy] < prices[i]) profit = prices[i] - prices[buy]; else{ buy = i; profit = -1; } max = std::max(max, profit); } return max; } };
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