LeetCode Search in Rotated Sorted Array II
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题目:
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.
题意:
继续上一题,如果有重复元素出现,那么该如何处理?还是如同上一题那么处理,采用二分搜索,然后考虑出现了重复的,那么就采用顺序遍历,知道找到了,否则就返回false.
public static boolean search(int[] nums, int target) { if(nums.length == 0 || nums == null) return false; int low = 0, high = nums.length - 1, mid; while(low <= high) { mid = low + (high - low) / 2; if(nums[mid] == target) return true; else if(nums[mid] > nums[low]) { if(target >= nums[low] && target < nums[mid]) high = mid - 1; else low = mid + 1; } else if(nums[mid] < nums[low]) { if(target > nums[mid] && target <= nums[high]) low = mid + 1; else high = mid - 1; } else //如果是相等的情况,也就是说又重复,那么考虑顺序遍历 { low++; } } return false; }
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