Search in Rotated Sorted Array_Leetcode_#33

1 题目

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.

2 解法

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

int nRotate = nLeft;
nLeft = 0;
nRight = nSize - 1;
while(nLeft <= nRight)
{
int nMid = nLeft + (nRight - nLeft) / 2;
int nMid2 = (nMid + nRotate) % nSize;
if(nums[nMid2] == target)
return nMid2;
else if(nums[nMid2] < target)
nLeft = nMid + 1;
else
nRight = nMid - 1;
}
return -1;
}
};