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This is combination of II and III.
1. when k >= length, it is totally II.
2. when k < length, it is another III. A localMaximum and globalMaximum subarray need to be maintained. But we can reduce the two dimensional DP to one dimensional DP
localMax[i][j] = max(local[i-1][j] + diff, globalMax[i-1][j-1] + max(0, diff)) ==> localMax[j] = max(localMax[j] + diff, globalMax[j-1] + max(0, diff))
globalMax[i][j] = max(globalMax[i-1][j], localMax[i][j]) ==> globalMax[j] = max(globalMax[j], localMax[j])
For here ,diff = prices[i+1] - prices[i]. max(0, diff) means only when the profit is positive, we add it to globalMax.
1 class Solution { 2 public: 3 int maxProfit(int k, vector<int> &prices) { 4 int len = prices.size(), result = 0; 5 if (len < 2) return 0; 6 if (len <= k) { 7 for (int i = 0; i < len-1; i++) { 8 if (prices[i+1] - prices[i] > 0) { 9 result += prices[i+1] - prices[i]; 10 } 11 } 12 } else { 13 vector<int> lMax(k+1, 0), gMax(k+1, 0); 14 for (int i = 0; i < len-1; i++) { 15 int diff = prices[i+1] - prices[i]; 16 for (int j = k; j >= 1; j--) { 17 lMax[j] = max(gMax[j-1] + max(0, diff), lMax[j] + diff); 18 gMax[j] = max(lMax[j], gMax[j]); 19 } 20 } 21 result = gMax[k]; 22 } 23 return result; 24 } 25 };
LeetCode – Refresh – Best Time to Buy and Sell Stock iv
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原文地址：http://www.cnblogs.com/shuashuashua/p/4346166.html
