[leetcode] 33. Search in Rotated Sorted Array 解题报告

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

Suppose a sorted array 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.

思路：做一次二分，分析应该搜索左边还是右边。每次二分有三种情况：

1. nums[mid] = target，则可以返回mid

2. nums[mid] < nums[right]，说明在[mid, right]区间是右边递增的区间，然后判断target是否在这个区间内

1）如果nums[mid] < target <= nums[right]，说明target在右边区间里，则left = mid + 1;

2）否则在左边区间里，搜索左边区间，right = mid - 1;

3. nums[mid] >= nums[right]，说明[elft, mid]区间是在左边的递增区间，然后判断target是否在这个左边区间里

1）如果nums[left] <= target < nums[mid]，说明target在这个区间里，则使right = mid - 1;

2）否则说明target在[mid, right]的不规则区间里，搜索右边区间，则使left = mid + 1;

然后考虑边界条件：

如果数组长度为1，则left = right，mid = left=right，如果nums[mid] = target，则直接返回，否则会进入条件3，然后循环结束，返回-1；

代码如下：

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

参考：http://bangbingsyb.blogspot.com/2014/11/leetcode-search-in-rotated-sorted-array.html