标签：blog os io for ar art div log
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.
排序的数组查找可以采用二分查找的方法~，有序数组被转动后依然可以采用二分的思想，但是要考虑到start，end与mid的大小决定递归调用二分还是自身。
因为分治后的数组一半严格有序，一半有序的转动数组。
1st （7 tries）
class Solution
{
public:
int search(int A[], int n, int target)
{
// Note: The Solution object is instantiated only once and is reused by each test case.
if(A[0]<A[n-1])
return orderfind(A,0,n-1,target);
else
return find(A,0,n-1,target);
}
int find(int A[],int start,int end,int target)
{
if(start > end||end < start)
{
return -1;
}
int mid = (start + end)/2;
if(target == A[mid])
{
return mid;
}
else if(target > A[mid])
{
if(A[mid] > A[start])
return find(A,mid+1,end,target);
else if(A[mid] == A[start])
return find(A,mid+1,end,target);
else
{
if(target == A[start])
return start;
else if(target > A[start])
return find(A,start,mid-1,target);
else
return orderfind(A,mid+1,end,target);
}
}
else
{
if(A[mid] < A[start])
return find(A,start,mid-1,target);
else if(A[mid] == A[start])
return find(A,mid+1,end,target);
else
{
if(target == A[start])
return start;
else if(target > A[start])
return orderfind(A,start,mid-1,target);
else
return find(A,mid+1,end,target);
}
}
}
int orderfind(int A[],int start,int end,int target)
{
if(start > end||end < start)
{
return -1;
}
int mid = (start + end)/2;
if(target == A[mid])
{
return mid;
}
else if(target > A[mid])
{
return orderfind(A,mid+1,end,target);
}
else
{
return orderfind(A,start,mid-1,target);
}
}
};
2nd （2 tries）
class Solution {
public:
int search(int A[], int n, int target) {
return search(A,0,n-1,target);
}
int search(int A[], int start, int end, int target) {
if(start > end)
return -1;
int mid = start + (end - start)/2;
//rotate n
if(A[start] <= A[end])
return bsearch(A,start,end,target);
if(A[mid] == target)
return mid;
else if(A[mid] > target) {
if(A[mid] >= A[start]) {
if(A[start] > target)
return search(A,mid+1,end,target);
else
return bsearch(A,start,mid-1,target);
}
else {
return search(A,start,mid-1,target);
}
}
else {
if(A[mid] >= A[start]) {
return search(A,mid+1,end,target);
}
else {
if(target > A[end]) {
return search(A,start,mid-1,target);
}
else {
return bsearch(A,mid+1,end,target);
}
}
}
}
int bsearch(int A[], int start, int end,int target) {
while(start <= end) {
int mid = start + (end - start)/2;
if(A[mid] == target)
return mid;
else if(A[mid] > target)
end = mid-1;
else
start = mid+1;
}
return -1;
}
};
【Leetcode】Search in Rotated Sorted Array,布布扣,bubuko.com
【Leetcode】Search in Rotated Sorted Array
标签：blog os io for ar art div log
原文地址：http://www.cnblogs.com/weixliu/p/3924324.html
