Binary Search

Author: gzwl

Find Minimum in Rotated Sorted Array II/Find Minimum in Rotated Sorted Array II.cpp

@@ -0,0 +1,25 @@
+/*
+    二分法，假设当期区间为[lhs,rhs),如果nums[lhs] < nums[rhs-1],说明是单调的,就返回nums[lhs]
+    否则说明不是单调的，就继续二分搜索[lhs，rhs)
+    由于多了相等的情况，每次搜索不一定能减去一半的区间
+    所以平均复杂度是O(logn)的，最坏复杂度是O(n)的
+
+*/
+class Solution {
+public:
+    int findMin(vector<int> &nums) {
+        return solve(nums,0,nums.size());
+    }
+    int solve(vector<int> &nums,int lhs,int rhs)
+    {
+        if(rhs - lhs == 1)    return nums[lhs];
+        else{
+            if(nums[lhs] < nums[rhs-1])   return nums[lhs];
+            else{
+                return min(solve(nums,lhs,(lhs + rhs) / 2),solve(nums,(lhs + rhs) / 2,rhs));
+            }
+        }
+    }
+};
+
+

Find Minimum in Rotated Sorted Array/Find Minimum in Rotated Sorted Array.cpp

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+/*
+    二分法，假设当期区间为[lhs,rhs),如果nums[lhs] < nums[rhs-1],说明是单调的,就返回nums[lhs]
+    否则说明不是单调的，就继续二分搜索[lhs，rhs)
+    最后的时间复杂度是O(logn)的
+
+*/
+class Solution {
+public:
+    int findMin(vector<int> &nums) {
+        return solve(nums,0,nums.size());
+    }
+    int solve(vector<int> &nums,int lhs,int rhs)
+    {
+        if(rhs - lhs == 1)    return nums[lhs];
+        else{
+            if(nums[lhs] < nums[rhs-1])   return nums[lhs];
+            else{
+                return min(solve(nums,lhs,(lhs + rhs) / 2),solve(nums,(lhs + rhs) / 2,rhs));
+            }
+        }
+    }
+};
+