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Search in Rotated Sorted Array II

描述

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

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

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

分析

允许重复元素,则上一题中如果A[left] <= A[mid],那么[left,mid]为递增序列的假设就不能成立了,比如[1,3,1,1,1]

既然A[left] <= A[mid]不能确定递增,那就把它拆分成两个条件:

  • A[left] < A[mid],则区间[left,mid]一定递增
  • A[left] == A[mid] 确定不了,那就left++,往下看一步即可。

代码

// Search in Rotated Sorted Array II
// Time Complexity: O(n),Space Complexity: O(1)
public class Solution {
public boolean search(int[] nums, int target) {
int first = 0, last = nums.length;
while (first != last) {
final int mid = first + (last - first) / 2;
if (nums[mid] == target)
return true;
if (nums[first] < nums[mid]) {
if (nums[first] <= target && target < nums[mid])
last = mid;
else
first = mid + 1;
} else if (nums[first] > nums[mid]) {
if (nums[mid] < target && target <= nums[last-1])
first = mid + 1;
else
last = mid;
} else
//skip duplicate one
first++;
}
return false;
}
};

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