Leetcode:best time to Buy and Sell Stock III problem Solving report

best time to Buy and Sell Stock III
Question Solution
Say you had an array for which the ith element was the price of a given stock on day I.

Design an algorithm to find the maximum profit. You are in most of the transactions.

Note:
Engage in multiple transactions on the same time (ie, you must sell the stock before you buy again).

Answer:
1. Scan from left to right to calculate the maximum profit for this interval of 0-i. Method can be found in stock first question
2. Scan from right to left to calculate the maximum profit for this interval of I-len. method as above.
3. Sweep through each node, plus the profit on the left and right. Record the maximum value.

Copy a little bit of someone else's explanation:

O (n^2) algorithm is easy to think of:

Find a Point J, the original price[0..n-1] divided into PRICE[0..J] and price[j. N-1], respectively, to find the maximum profit of two paragraphs.

To optimize:

For the point j+1, price[0..j+1] maximum profit, a lot of work is repeated, in the PRICE[0..J] the largest profit has been done.

Similar to the best time to Buy and Sell Stock, the maximum profit of price[0..j+1] can be introduced at O (1) from PRICE[0..J].

But how to get from Price[j. N-1] launched Price[j+1..n-1]? In turn thinking, we can use O (1) time by price[j+1..n-1] to launch price[j. N-1].

The final algorithm:

Array L[i] records the maximum profit of PRICE[0..I],

Array R[i] records the price[i. N] the maximum profit.

Known L[i], seeking l[i+1] is simple, also known r[i], R[i-1] is also very easy.

Finally, we use O (n) time to find the largest l[i]+r[i], which is the problem. (The final step can be combined in the second step).

ref:http://blog.csdn.net/pickless/article/details/12034365

Code Listing 1:

1  Public intMaxprofit (int[] prices) {2         if(Prices = =NULL|| Prices.length = = 0) {3             return0;4         }5         6         intLen =prices.length;7         int[] left =New int[Len];8         int[] right =New int[Len];9         Ten         intMin = prices[0]; OneLeft[0] = 0; A          for(inti = 1; i < Len; i++) { -Min =math.min (min, prices[i]); -Left[i] = Math.max (Left[i-1], prices[i]-min); the         } -          -         intmax = Prices[len-1]; -Right[len-1] = 0; +          for(inti = len-2; I >= 0; i--) { -Max =Math.max (Max, prices[i]); +Right[i] = Math.max (right[i + 1], max-prices[i]); A         } at          -         intRST = 0; -          for(inti = 0; i < Len; i++) { -rst = Math.max (RST, left[i] +right[i]); -         } -          in         returnrst; -}

View Code

Code Listing 2:

1  Public classSolution {2      Public intMaxprofit (int[] prices) {3         if(Prices = =NULL) {4             return0;5         }6         7         intRET = 0;8         9         intLen =prices.length;Ten         int[] Leftprofile =New int[Len]; One         intProfile = 0; A          -         intMin =Integer.max_value; -          for(inti = 0; i < Len; i++) { theMin =math.min (min, prices[i]); -Profile = Math.max (profile, Prices[i]-min); -Leftprofile[i] =Profile ; -         } +          -         intMax =Integer.min_value; +Profile = 0; A          for(inti = len-1; I >= 0; i--) { atMax =Math.max (Max, prices[i]); -Profile = Math.max (profile, Max-prices[i]); -              -             //Sum the left profit and the right profit. -ret = Math.max (ret, profile +leftprofile[i]); -         } in          -         returnret; to     } +}

View Code

DP idea:

1 //DP Solution:2      Public intMaxprofit (int[] prices) {3         if(Prices = =NULL|| Prices.length = = 0) {4             return0;5         }6         7         intRET = 0;8         9         intLen =prices.length;Ten         int[] Leftprofile =New int[Len]; One          A         intMin = prices[0]; -Leftprofile[0] = 0; -          for(inti = 1; i < Len; i++) { theMin =math.min (min, prices[i]); -Leftprofile[i] = Math.max (Leftprofile[i-1], prices[i]-min); -         } -          +         intMax =Integer.min_value; -         intProfile = 0; +          for(inti = len-1; I >= 0; i--) { AMax =Math.max (Max, prices[i]); atProfile = Math.max (profile, Max-prices[i]); -              -             //Sum the left profit and the right profit. -ret = Math.max (ret, profile +leftprofile[i]); -         } -          in         returnret; -}

View Code

GitHub Code Links

Leetcode:best time to Buy and Sell Stock III problem Solving report