81. Search in Rotated Sorted Array II

最新推荐文章于 2023-11-12 22:36:10 发布

OovEver

最新推荐文章于 2023-11-12 22:36:10 发布

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分类专栏： LeetCode 文章标签： search array sort leetcode-java

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本文介绍了一种在旋转过的有序数组中查找特定元素的方法。通过改进的二分查找算法实现高效搜索，即使数组中存在重复元素也能正确处理。

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81. Search in Rotated Sorted Array II

Suppose an array sorted in ascending order 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).

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

The array may contain duplicates.

题目大意：给定一个旋转后的数组，写一个函数，判断给定的数，是否在数组中。
思路：
1. 直接遍历数组，判断是否有target，时间复杂度O（N）。
2. 二分查找，查找目标数，时间复杂度O（logN）.

代码：

package Array;



/**
* @Author OovEver
* @Date 2017/11/28 10:10
*/
public class Solution {
  public boolean search(int[] nums, int target) {
      int start = 0, end = nums.length - 1, mid = -1;
      while(start <= end) {
          mid = (start + end) / 2;
          if (nums[mid] == target) {
              return true;
          }
          //If we know for sure right side is sorted or left side is unsorted
          if (nums[mid] < nums[end] || nums[mid] < nums[start]) {
              if (target > nums[mid] && target <= nums[end]) {
                  start = mid + 1;
              } else {
                  end = mid - 1;
              }
              //If we know for sure left side is sorted or right side is unsorted
          } else if (nums[mid] > nums[start] || nums[mid] > nums[end]) {
              if (target < nums[mid] && target >= nums[start]) {
                  end = mid - 1;
              } else {
                  start = mid + 1;
              }
              //If we get here, that means nums[start] == nums[mid] == nums[end], then shifting out
              //any of the two sides won't change the result but can help remove duplicate from
              //consideration, here we just use end-- but left++ works too
          } else {
//                low++也可以
              end--;
          }
      }

      return false;
  }

}