Leetcode(81) Search in Rotated Sorted Array II

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Description

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

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Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true

Example 2:

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Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

Follow up:

  • This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
  • Would this affect the run-time complexity? How and why?

解法

这是之前那道在有序翻转数组中查找的加强版，唯一不同的是这里可能出现重复值，之前那题的方法为：如果中间的数小于最右边的数，则右半段是有序的，若中间数大于最右边数，则左半段是有序的。而如果可以有重复值，就会出现来面两种情况，[3 1 1] 和 [1 1 3 1]，对于这两种情况中间值等于最右值时，目标值3既可以在左边又可以在右边，那怎么办么，对于这种情况其实处理非常简单，只要把最右值向左一位即可继续循环，如果还相同则继续移，直到移到不同值为止。其他的与之前那题完全一致。

具体代码如下：

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