Best Time to Buy and Sell Stock III -- LeetCode
2016-10-12 20:45
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原题链接: http://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/
因为如果diff大于0,且第i-1天也有交易,那么相当于i天和i-1天的交易合并了,相当于j-1次交易。并不是题目要求最多j次交易。
那么
最终结果就是 global
[k];
我们知道,动规所用的二维辅助数组可以降为一维的,即只用大小为k的一维数组记录到达第i天时的局部最优解和全局最优解。需要注意的是,由于第i天时交易k次的最优解依赖于第i-1天时交易k-1次的最优解,所以数组更新应当从后往前(即从k到1)更新。
为什么不能用 dp[i][j] = max(dp[i-1][j-1], dp[i-1][j] + diff)
参考:http://liangjiabin.com/blog/2015/04/leetcode-best-time-to-buy-and-sell-stock.html因为如果diff大于0,且第i-1天也有交易,那么相当于i天和i-1天的交易合并了,相当于j-1次交易。并不是题目要求最多j次交易。
local[i][j]表示第i天一定卖出的最大利润。global[i][j]表示到第i天最多j次交易的最大利润。
参考 http://blog.csdn.net/linhuanmars/article/details/23236995那么
global[i][j] = max(local[i][j] + global[i - 1][j]);//最后一天交易 和 最后一天不交易 相比较 local[i][j] = max(local[i-1][j]+diff, global[i-1][j-1]+max(diff, 0));
local[i-1][j]+diff 表示 在第i-1天一定有交易 且是第j次交易。
global[i-1][j-1]+max(diff, 0) 表示 i-1天或之前进行j - 1次交易 的最大利润 加上 最后一天交易的。
最终结果就是 global
[k];
我们知道,动规所用的二维辅助数组可以降为一维的,即只用大小为k的一维数组记录到达第i天时的局部最优解和全局最优解。需要注意的是,由于第i天时交易k次的最优解依赖于第i-1天时交易k-1次的最优解,所以数组更新应当从后往前(即从k到1)更新。
class Solution { public: int maxProfit(int k, vector<int>& prices) { int n = prices.size(); if(n <= 1)return 0; if(k >= n)return maxProfit2(prices); //global[i][j]=max(local[i][j],global[i-1][j]) //local[i][j]=max(global[i-1][j-1]+max(diff,0),local[i-1][j]+diff) int local[k+1] = {0}; int global[k+1] = {0}; for(int i = 1; i < n; i++) { int diff = prices[i] - prices[i - 1]; for(int j = k; j > 0; j--) { local[j] = max(global[j - 1] + max(diff, 0), local[j] + diff); global[j] = max(local[j], global[j]); } } return global[k]; } int maxProfit2(vector<int>& prices){ int ans = 0; for(int i = 1; i < prices.size(); i++) { if(prices[i] > prices[i - 1]) ans += prices[i] - prices[i - 1]; } return ans; } };
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