update

Author: FeiCoding

(#81)SearchinRotatedSortedArrayII/SearchinRotatedSortedArrayII.java

@@ -1,38 +1,26 @@
-public boolean search(int[] nums, int target) {
-    int start  = 0, end = nums.length - 1;
-    
-    //check each num so we will check start == end
-    //We always get a sorted part and a half part
-    //we can check sorted part to decide where to go next
-    while(start <= end){
-        int mid = start + (end - start)/2;
-        if(nums[mid] == target) return true;
-        
-        //if left part is sorted
-        if(nums[start] < nums[mid]){
-            if(target < nums[start] || target > nums[mid]){
-                //target is in rotated part
-                start = mid + 1;
+class Solution {
+    public boolean search(int[] nums, int target) {
+        int start = 0;
+        int end = nums.length - 1;
+        while(start <= end){
+            int mid = (start + end) / 2;
+            if(nums[mid] == target)
+                return true;
+            if(nums[mid] > nums[start]){
+                if(target < nums[mid] && target >= nums[start])
+                    end = mid - 1;
+                else
+                    start = mid + 1;
+            }else if(nums[mid] < nums[start]){
+                if(nums[mid] < target && target <= nums[end])
+                    start = mid + 1;
+                else
+                    end = mid - 1;
+                
             }else{
-                end = mid - 1;
+                start ++;
             }
-        }else if(nums[start] > nums[mid]){
-            //right part is rotated
-            
-            //target is in rotated part
-            if(target < nums[mid] || target > nums[end]){
-                end = mid -1;
-            }else{
-                start = mid + 1;
-            }
-        }else{
-            //duplicates, we know nums[mid] != target, so nums[start] != target
-            //based on current information, we can only move left pointer to skip one cell
-            //thus in the worest case, we would have target: 2, and array like 11111111, then
-            //the running time would be O(n)
-            start ++;
         }
+        return false;
     }
-    
-    return false;
 }