update 81

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_81.java

@@ -1,44 +1,66 @@
 package com.fishercoder.solutions;
 
+/**
+ * 81. Search in Rotated Sorted Array II
+ * <p>
+ * There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).
+ * Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is
+ * [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].
+ * Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.
+ * You must decrease the overall operation steps as much as possible.
+ * <p>
+ * Example 1:
+ * Input: nums = [2,5,6,0,0,1,2], target = 0
+ * Output: true
+ * <p>
+ * Example 2:
+ * Input: nums = [2,5,6,0,0,1,2], target = 3
+ * Output: false
+ * <p>
+ * Constraints:
+ * 1 <= nums.length <= 5000
+ * -104 <= nums[i] <= 104
+ * nums is guaranteed to be rotated at some pivot.
+ * -104 <= target <= 104
+ * Follow up: This problem is similar to Search in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?
+ */
 public class _81 {
-
     public static class Solution1 {
         public boolean search(int[] nums, int target) {
-            int start = 0;
-            int end = nums.length - 1;
+            int left = 0;
+            int right = nums.length - 1;
 
-            //check each num so we will check start == end
+            //check each num so we will check left == right
             //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;
+            while (left <= right) {
+                int mid = left + (right - left) / 2;
                 if (nums[mid] == target) {
                     return true;
                 }
 
-                //if left part is sorted
-                if (nums[start] < nums[mid]) {
-                    if (target < nums[start] || target > nums[mid]) {
+                if (nums[left] < nums[mid]) {
+                    //if left part is sorted
+                    if (target < nums[left] || target > nums[mid]) {
                         //target is in rotated part
-                        start = mid + 1;
+                        left = mid + 1;
                     } else {
-                        end = mid - 1;
+                        right = mid - 1;
                     }
-                } 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 if (nums[left] > nums[mid]) {
+                    //right part is sorted
+                    if (target < nums[mid] || target > nums[right]) {
+                        //target is in rotated part
+                        right = mid - 1;
                     } else {
-                        start = mid + 1;
+                        left = mid + 1;
                     }
                 } else {
-                    //duplicates, we know nums[mid] != target, so nums[start] != target
+                    //duplicates, we know nums[mid] != target, so nums[left] != target
                     //based on current information, we can only move left pointer to skip one cell
                     //thus in the worst case, we would have target: 2, and array like 11111111, then
                     //the running time would be O(n)
-                    start++;
+                    left++;
                 }
             }
             return false;