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leetcode之Search in Rotated Sorted Array，剑指offer之旋转数组的最小数字
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.
class Solution {
public:
int search(int A[], int n, int target) {
int left = 0,right = n-1;
while(left <= right)
{
int mid = left + ((right-left)>>1);
if(A[mid] == target)return mid;
if(A[mid] > A[right])//第一段递增区间
{
if(A[left] <= target && target < A[mid])right = mid-1;
else left = mid + 1;
}
else //第二段递增区间
{
if(A[mid] < target && target <= A[right])left = mid+1;
else right = mid - 1;
}
}
return -1;
}
};Search in Rotated Sorted Array II
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.
class Solution {
public:
bool search(int A[], int n, int target) {
int left = 0,right = n-1;
while(left <= right)
{
int mid = left + ((right-left)>>1);
if(A[mid] == target)return true;
if(A[mid] == A[right])//遍历最后一段
{
for(;right >= mid;--right)
{
if(A[right] == target)return true;
}
}
else if(A[mid] > A[right])//第一段递增区间
{
if(A[left] <= target && target < A[mid])right = mid-1;
else left = mid + 1;
}
else //第二段递增区间
{
if(A[mid] < target && target <= A[right])left = mid + 1;
else right = mid - 1;
}
}
return false;
}
};剑指offer：把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个递增排序的数组的一个旋转，输出旋转数组的最小元素。例如数组{3,4,5,1,2}为{1,2,3,4,5}的一个旋转，该数组的最小值为1。
思路：这里给出和上面两个类似的思路，和剑指offer上的不太一样，我觉得剑指offer上的不好理解，但是这个算法的效率没有书上的高，九度空间上超时了。
int RotateArray(int* data,int n)
{
int left = 0,right = n-1,minValue = http://www.mamicode.com/INT_MAX;>
leetcode之Search in Rotated Sorted Array，剑指offer之旋转数组的最小数字
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