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
1 public int search(int[] A, int target) { 2 int first = 0, last = A.length; 3 while (first != last) { 4 int mid = (first + last) / 2; 5 if (A[mid] == target) 6 return mid; 7 if (A[first] <= A[mid]) { 8 if (A[first] <= target && target < A[mid]) 9 last = mid; 10 else 11 first = mid + 1; 12 } else { 13 if (A[mid] < target && target <= A[last - 1]) 14 first = mid + 1; 15 else 16 last = mid; 17 } 18 } 19 return -1; 20 }
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。
1 public boolean search(int[] A, int target) { 2 int first = 0, last = A.length; 3 while (first != last) { 4 int mid = (first + last) / 2; 5 if (A[mid] == target) 6 return true; 7 if (A[first] < A[mid]) { 8 if (A[first] <= target && target < A[mid]) 9 last = mid; 10 else 11 first = mid + 1; 12 } else if (A[first] > A[mid]) { 13 if (A[mid] < target && target <= A[last - 1]) 14 first = mid + 1; 15 else 16 last = mid; 17 } else { 18 first++; 19 } 20 } 21 return false; 22 }