Search in Rotated Sorted Array I II

Search in Rotated Sorted Array

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).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

用二分查找去做

 public int search(int[] A, int target) {
         int first = 0, last = A.length;
         while (first != last) {
             int mid = (first + last) / 2;
             if (A[mid] == target)
                 return mid;
             if (A[first] <= A[mid]) {
                 if (A[first] <= target && target < A[mid])
                     last = mid;
                 else
                     first = mid + 1;
             } else {
                 if (A[mid] < target && target <= A[last - 1])
                     first = mid + 1;
                 else
                     last = mid;
             }
         }
         return -1;
     }

Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

允许重复，造成了如果A[m]>=A[l], 那么[l,m] 为递增序列的假设就不能成立了，比如[1,3,1,1,1]。

如果A[m]>=A[l] 不能确定递增，那就把它拆分成两个条件：
• 若A[m]>A[l]，则区间[l,m] 一定递增
• 若A[m]==A[l] 确定不了，那就l++，往下看一步即可。

因此复杂度从O(lgn)变成了O(n)。具体说明参见剑指offer上查找旋转数组的最小元素。

 public boolean search(int[] A, int target) {
         int first = 0, last = A.length;
         while (first != last) {
             int mid = (first + last) / 2;
             if (A[mid] == target)
                 return true;
             if (A[first] < A[mid]) {
                 if (A[first] <= target && target < A[mid])
                     last = mid;
                 else
                     first = mid + 1;
             } else if (A[first] > A[mid]) {
                 if (A[mid] < target && target <= A[last - 1])
                     first = mid + 1;
                 else
                     last = mid;
             } else {
                 first++;
             }
         }
         return false;
     }