81. Search in Rotated Sorted Array II
摘要：
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,...
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:
Input: nums = [2,5,6,0,0,1,2], target = 0 Output: true
Example 2:
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?
難度：medium
題目：假設一個按升序排序的陣列在某個未知的軸上旋轉。搜尋給定目標值，如果找到返回true，否則返回false。
思路：二叉搜尋。
分4種情況。
case 1：從頭到尾升序。 case 2：從頭到尾降序。 case 3：升序的元素多，降序少。 case 4：升序的元素少，降序多。
Runtime: 0 ms, faster than 100.00% of Java online submissions for Search in Rotated Sorted Array II.
Memory Usage: 37.6 MB, less than 1.00% of Java online submissions for Search in Rotated Sorted Array II.
class Solution {
public boolean search(int[] nums, int target) {
int left = 0, right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (target == nums[mid]) {
return true;
}
// left < mid </> right
if (nums[left] < nums[mid]) {
if (target > nums[mid] || target < nums[left]) {
left = mid + 1;
} else {
right = mid - 1;
}
} else if (nums[left] > nums[mid]) { // left > mid </> right
if (target < nums[mid] || target > nums[right]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else { // left = mid
left += 1;
}
}
return false;
}
}