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.
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两种思路：
1（未思考：直接做，仔细考虑各种情况来分析）
package leetcode;
public class SearchinRotatedSortedArray {
public static int search(int[] nums, int target) {
int index = -1;
if (nums.length < 1 || nums == null)
return -1;
if (nums.length == 1) {
if (target == nums[0])
return 0;
return -1;
}
if (target == nums[0]) {
index = 0;
} else if (target == nums[nums.length - 1]) {
index = nums.length - 1;
}
if(nums[0]<nums[nums.length-1]){
index = Erfen(target,nums);
}else{
if (target < nums[0] && target > nums[nums.length - 1]) {
index = -1;
} else if (target > nums[0]) {
int i = 0;
while(nums[i]<=nums[i+1] && i < nums.length - 1){
if(nums[i]==target){
index = i;
break;
}
i++;
}if(target == nums[i]){
index=i;
}
} else if (target < nums[nums.length - 1]) {
int j = nums.length-1;
while(nums[j-1]< nums[j] && j > 0){
if(nums[j]==target){
index = j;
break;
}
j--;
}
if(target == nums[j]){
index=j;
}
}
}
return index;
}
public static int Erfen(int target, int[] array) {
int left = 0;
int right = array.length - 1;
while (left < right-1 ) {
int mid = left + (right - left) / 2;
if(target >= array[mid]){
left = mid;
}else{
right = mid;
}
}
if(target == array[left]){
return left;
}else if(target == array[right]){
return right;
}else {
return -1;
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] array = { 3,4,5,6,7,8,9 };
System.out.println(search(array,5));
}
}
2（借鉴网络，思考后，在数组上设定左右标，利用二分法+递归）
思路：在整个数组中，先二分，找到有序的那一边然后在有序的一边判断是否含有target；如果没有，继续在无序的一半二分，在有序的一边查找
package leetcode;
public class newSearchinRotatedSortedArray {
public static int search(int[] nums, int target) {
return search(nums,0,nums.length-1,target);
}
public static int search(int[] nums, int left, int right, int target) {
// TODO Auto-generated method stub
if(left > right) return -1;
int mid = (left+right)/2;
//78123456
if(nums[mid]==target){
return mid;
}
if(nums[left]<=nums[mid]){
if(target<=nums[mid]&&target>=nums[left]) {
return search(nums,left,mid-1,target); //二分法加上“=”号！当使用二分法时，可以在每个判断大小的地方加“=”？
}else{
return search(nums,mid+1,right,target);
}
}else if(nums[mid]<=nums[nums.length-1]){
if(target>=nums[mid]&&target<=nums[right]){
return search(nums,mid+1,right,target);
}else{
return search(nums,left,mid-1,target);
}
}
return -1;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] array = { 7,8,9,3,4,5,6 };
System.out.println(search(array,3));
}
}