【Leetcode】Search in Rotated Sorted Array II
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题目链接:https://leetcode.com/problems/search-in-rotated-sorted-array-ii/
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.
思路:
找到逆序的位置,然后根据该位置将数组分为两半,分别二分查找
算法:
public boolean search(int[] nums, int target) { int index = 0; // 找到旋转点即逆序的点 for (int i = 0; i < nums.length - 1; i++) { if (nums[i] > nums[i + 1]) { index = i; } } boolean left_res = search(nums, target, 0, index); boolean right_res = search(nums, target, index + 1, nums.length - 1); return left_res || right_res; } public boolean search(int nums[], int target, int start, int end) { int left = start, right = end, mid = 0; while (left <= right) { mid = left + (right - left) / 2; if (nums[mid] == target) { return true; } else if (nums[mid] > target) { right = mid - 1; } else { left = mid + 1; } } return false; }
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