Add 'Find Minimum in Rotated Sorte Array II' solution.

Author: liuyingshan

3Sum.cpp

@@ -1,4 +1,6 @@
 /*
+  3Sum
+
   Given an array S of n integers, are there elements a, b, c in S such
   that a + b + c = 0? Find all unique triplets in the array which gives
   the sum of zero.

Find_Minimum_in_Rotated_Sorted_Array_II.cpp

@@ -0,0 +1,35 @@
+/*
+  Find Minimum in Rotated Sorted Array II
+
+  Suppose a sorted array is rotated at some pivot unknown to you beforehand.
+  (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
+  Find the minimum element.The array may contain duplicates.
+*/
+
+class Solution {
+public:
+    int findSubMin(vector<int> &num, int l, int r) {
+        if (l > r) return LONG_MAX;
+        if (l == r) return num[l];
+
+        // Enumerate all relations of num[l] and num[mid], num[mid] and num[r].
+        // {num[l]=num[mid], num[l]<num[mid], num[l]>num[mid]} x {num[mid]=num[r], num[mid]<num[r], num[mid]>num[r]}
+        // It has 9 conditions, which are redundant.
+        // Not to merge synonymic statements in order to improve readability.
+        int mid = (l + r) / 2;
+        if (num[l] == num[mid] && num[mid] == num[r])
+            return std::min(num[l], std::min(findSubMin(num, l+1, mid-1), findSubMin(num, mid+1, r-1)));
+        if (num[l] == num[mid] && num[mid] < num[r]) return num[l];
+        if (num[l] == num[mid] && num[mid] > num[r]) return findSubMin(num, mid+1, r);
+        if (num[l] < num[mid] && num[mid] == num[r]) return num[l];
+        if (num[l] < num[mid] && num[mid] < num[r]) return num[l];
+        if (num[l] < num[mid] && num[mid] > num[r]) return findSubMin(num, mid+1, r);
+        if (num[l] > num[mid] && num[mid] == num[r]) return findSubMin(num, l+1, mid);
+        if (num[l] > num[mid] && num[mid] < num[r]) return findSubMin(num, l+1, mid);
+        if (num[l] > num[mid] && num[mid] > num[r]) return findSubMin(num, mid+1, r);
+    }
+
+    int findMin(vector<int> &num) {
+        return findSubMin(num, 0, num.size()-1);
+    }
+};