LeetCode 81. Search in Rotated Sorted Array II

原创于 2017-07-14 00:09:20 发布 · 211 阅读

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针对含有重复元素的旋转排序数组，实现一个函数判断目标值是否存在于数组中。本篇文章介绍了一种二分查找的改进算法，该算法能够有效处理含有重复元素的情况。

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Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

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.

比起之前的题多了重复数字，需要做的是，如果相同则把左边界前移

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