33.81.Search in Rotated Sorted Array-优快云博客

本文介绍了一种在旋转排序数组中查找特定目标值的算法。利用二分法思想，该算法能在O(logN)的时间复杂度内高效解决问题，并讨论了允许重复元素情况下的实现细节。

Suppose a sorted array 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).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

思路: 二分法, 将元素与中间元素进行对比, 中间元素要么比左边的元素大,要么就比右边的元素小,因此可以分成2种情况进行分析; 每一种情况又分为2种.从左边到中间要么是递增的,要么不是.

时间复杂度: O(logN)

public int search(int[] nums, int target) {
        
        if(nums==null)
            return -1;
        int len=nums.length; 
        int mid,left=0,right=len-1;
        while(left<=right){
            mid=(left+right)/2;
            if(nums[mid]==target)
                return mid;
                
            if(nums[mid]>=nums[left]){
                if(target>=nums[left]&&target<nums[mid])
                    right=mid-1;
                else
                    left=mid+1;
            }
            
            if(nums[mid]<nums[right]){
                if(target>nums[mid]&&target<=nums[right])
                    left=mid+1;
                else
                    right=mid-1;
            }
                
        }
        return -1;
    }

81. Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array":
duplicates are allowed.

举例:

[1,3,1,1]

[1,3,1,1,1]

[1,1,1,3,1]

这种情况,mid和right和left的值都相同时,无法判断target属于哪一边,因此只能将left++,right--.

代码:

public class Solution {
    public boolean search(int[] nums, int target) {
        if(nums==null)
            return false;
        
        int n=nums.length;
        int left=0,right=n-1;
        int mid;
        while(left<=right){
            mid=left+(right-left)/2;
            if(nums[mid]==target)
                return true;
            if(nums[mid]==nums[left]&&nums[mid]==nums[right]){
                left++;
                right--;
            }
            else if(nums[mid]>=nums[left]){
                if(nums[mid]>target&&target>=nums[left])
                    right=mid-1;
                else
                    left=mid+1;
            }
                
            else if(nums[mid]<=nums[right]){
                if(target>nums[mid]&&target<=nums[right])
                    left=mid+1;
                else
                    right=mid-1;
            }
        }
            
        return false;
    }
}