[leetcode]81. Search in Rotated Sorted Array II(Java)
2017-07-02 10:41
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https://leetcode.com/problems/search-in-rotated-sorted-array-ii/#/description
Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e.,
Write a function to determine if a given target is in the array.
The array may contain duplicates.
Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e.,
0 1 2 4 5 6 7might become
4 5 6 7 0 1 2).
Write a function to determine if a given target is in the array.
The array may contain duplicates.
package go.jacob.day702; public class Demo1 { public boolean search(int[] nums, int target) { if (nums == null || nums.length < 1) return false; int left = 0, right = nums.length - 1; int mid; while (left <= right) { mid = left + (right - left) / 2; if (target == nums[mid]) { return true; } if (nums[mid] > nums[left]) { if (target < nums[mid] && target >= nums[left]) right = mid - 1; else left = mid + 1; } else if (nums[mid] < nums[left]) { if (target > nums[mid] && target <= nums[right]) left = mid + 1; else right = mid - 1; } else { // nums[left]==nums[mid] if (nums[right] > nums[mid] && target > nums[mid] && target <= nums[right]) left = mid + 1; else { //如果left,mid,right三个元素都相同,或者left,mid相同, //target不在[mid,right]递增区间里,只能循环遍历 for (int i = left; i <= right; i++) if (target == nums[i]) { return true; } break; } } } return false; } }
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