leetcode题解-33. Search in Rotated Sorted Array && 81. Search in Rotated Sorted Array II

31，题目：

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).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

这道题目是在一个反转中的数组中寻找一个数的索引，我们可以采用Binary Search的方法进行寻找，与传统二分法不同的是，这里需要考虑中间值是位于反转值之前还是之后，需要多加一个判断，代码如下所示：

public static int search(int[] nums, int target) {
int start = 0;
int end = nums.length - 1;
while (start <= end){
int mid = (start + end) / 2;
if (nums[mid] == target)
return mid;
//start小于mid说明mid在反转值之前
if (nums[start] <= nums[mid]){
if (target < nums[mid] && target >= nums[start])
end = mid - 1;
else
start = mid + 1;
}
//mid小于end说明mid在反转值之后
if (nums[mid] <= nums[end]){
if (target > nums[mid] && target <= nums[end])
start = mid + 1;
else
end = mid - 1;
}
}
return -1;
}

81，本体是基于33题，加了一个约束条件：数组中的数是可重复的，我们需要添加一个判断的就是前后元素是否相等，如果相等的话直接跳过即可。代码入下：

public boolean search(int[] nums, int target) {
int left=0, right=nums.length-1, mid;

while(left <= right){
mid = (left+right)/2;
if(nums[mid] == target)
return true;

if(nums[mid] < nums[left] || nums[mid] < nums[right]){
if(target > nums[mid] && target <= nums[right])
left = mid+1;
else
right = mid -1;
}else if(nums[mid] > nums[left] || nums[mid] > nums[right]){
if(target < nums[mid] && target >= nums[left])
right = mid-1;
else
left = mid + 1;
}else
right--;

}
return false;
}