From 4410ee28916cdc6b31f8c4614b291d60c62997f4 Mon Sep 17 00:00:00 2001
From: bluepp <bluepp@gmail.com>
Date: Sun, 13 Sep 2015 16:52:12 +0800
Subject: [PATCH] Create MaximumSubarrayII.h

---
 MaximumSubarrayII.h | 53 +++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 53 insertions(+)
 create mode 100644 MaximumSubarrayII.h

diff --git a/MaximumSubarrayII.h b/MaximumSubarrayII.h
new file mode 100644
index 0000000..86789ca
--- /dev/null
+++ b/MaximumSubarrayII.h
@@ -0,0 +1,53 @@
+/*
+  2015-09-13
+  bluepp
+  May the force be with me!
+ 
+LintCode
+Given an array of integers, find two non-overlapping subarrays which have the largest sum.
+
+The number in each subarray should be contiguous.
+
+Return the largest sum.
+
+Have you met this question in a real interview? Yes
+Example
+For given [1, 3, -1, 2, -1, 2], the two subarrays are [1, 3] and [2, -1, 2] or [1, 3, -1, 2] and [2],
+they both have the largest sum 7.
+http://www.lintcode.com/en/problem/maximum-subarray-ii/#
+*/
+
+    int maxTwoSubArrays(vector<int> nums) {
+        // write your code here
+        int n = nums.size();
+        
+        int sum1 = 0, maxsum1 = INT_MIN;
+        int dp1[n];
+        memset(dp1, 0, sizeof(dp1));
+        
+        for(int i = 0; i < n; i++)
+        {
+            sum1 = max(sum1+nums[i], nums[i]);
+            maxsum1 = max(maxsum1, sum1);
+            dp1[i] = maxsum1;
+        }
+        
+        int dp2[n];
+        memset(dp2, 0, sizeof(dp2));
+        dp2[n-1] = nums[n-1];;
+        int sum2 = 0, maxsum2 = INT_MIN;
+        for (int i = n-1; i >= 0; i--)
+        {
+            sum2 = max(nums[i], sum2+nums[i]);
+            maxsum2 = max(maxsum2, sum2);
+            dp2[i] = maxsum2;
+        }
+        
+        int ret = INT_MIN;
+        for (int i = 0; i < n-1; i++)
+        {
+            ret = max(ret, dp1[i] + dp2[i+1]);
+        }
+        
+        return ret;
+    }