LeetCode 81 Search in Rotated Sorted Array II (二分)

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.

题目链接：https://leetcode.com/problems/search-in-rotated-sorted-array-ii/

题目分析：和Search in Rotated Sorted Array本题允许重复元素的存在，其实只要修改一个地方，详见程序，但这个修改会使得程序在最坏的情况(比如所有元素都相同且无解)时，时间复杂度退化到O(n)

public class Solution {

public boolean search(int[] nums, int target) {
int l = 0, r = nums.length - 1, mid = 0;
while (l <= r) {
mid = (l + r) >> 1;
if (nums[mid] == target) {
return true;
}
//如果中点和当前左端点值相同，则不用考虑这个左端点了
if (nums[mid] == nums[l]) {
l ++;
}
else if (nums[mid] < nums[l]) {
if (target > nums[mid] && target <= nums[r]) {
l = mid + 1;
}
else {
r = mid - 1;
}
}
else {
if (target >= nums[l] && target < nums[mid]) {
r = mid - 1;
}
else {
l = mid + 1;
}
}
}
return false;
}
}