81. Search in Rotated Sorted Array II

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;  }}

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