Minor refactor.

BestTimetoBuyandSellStockIII.h

@@ -1,6 +1,7 @@
 /*
  Author:     Annie Kim, anniekim.pku@gmail.com
  Date:       Apr 28, 2013
+ Update:     Aug 22, 2013
  Problem:    Best Time to Buy and Sell Stock III
  Difficulty: Medium
  Source:     http://leetcode.com/onlinejudge#question_123
@@ -10,36 +11,37 @@
  Note:
  You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
 
- Solution: dp.
+ Solution: dp. max profit =  max { l2r[0...i] + r2l[i+1...N-1] }.
+                         0 <= i <= N-1
  */
 
 class Solution {
 public:
     int maxProfit(vector<int> &prices) {
         int N = prices.size();
-        vector<int> l2r(N, 0);   // [1, i]
-        vector<int> r2l(N, 0);   // [i + 1, N]
-
-        int imin = 0;
+        if (N <= 1) return 0;
+
+        int l2r[N], r2l[N];
+        l2r[0] = 0;
+        r2l[N-1] = 0;
+
+        int minn = prices[0];
         for (int i = 1; i < N; ++i)
         {
-            if (prices[i] < prices[imin])
-                imin = i;
-            l2r[i] = max(l2r[i-1], prices[i] - prices[imin]);
+            minn = min(minn, prices[i]);
+            l2r[i] = max(l2r[i-1], prices[i] - minn);
         }
-
-        int imax = N - 1;
-        for (int i = N - 3; i >= 0; --i)
+        
+        int maxx = prices[N-1];
+        for (int i = N-2; i >= 0; --i)
         {
-            if (prices[i+1] > prices[imax])
-                imax = i + 1;
-            r2l[i] = max(r2l[i+1], prices[imax] - prices[i+1]);
+            maxx = max(maxx, prices[i]);
+            r2l[i] = max(r2l[i+1], maxx - prices[i]);
         }
-
-        int res = 0;
-        for (int i = 0; i < N; ++i)
-            res = max(res, l2r[i] + r2l[i]);
+        
+        int res = l2r[N-1];
+        for (int i = 0; i < N-1; ++i)
+            res = max(res, l2r[i] + r2l[i+1]);
         return res;
     }
 };
-