leetcode — search-in-rotated-sorted-array

/**
* Source : https://oj.leetcode.com/problems/search-in-rotated-sorted-array/ *
* Created by lverpeng on 2017/7/13.
*
* 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.
*
*/
public class SearchInRotatedSortedArray {

/**
* 有序数组以某一个支点被翻转过，在数组中查找某一个元素
*
* 数组是局部有序的，使用二分查找的过程中使用这个特点
*
* @param num
* @return
*/
public int search (int[] num, int target) {
int left = 0;
int right = num.length - 1;
int mid = 0;
while (left <= right) {
mid = (left + right) / 2;
if (num[mid] == target) {
return mid;
}
// left- mid之间是局部有序的
if (num[left] <= num[mid]) {
if (num[left] <= target && target < num[mid]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else {
// mid - right是有序的
if (mid < target && target <= num[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
}
return -1;
}

public static void main(String[] args) {
SearchInRotatedSortedArray searchInRotatedSortedArray = new SearchInRotatedSortedArray();
int[] arr = new int[]{4, 5, 6, 7, 0, 1, 2};
System.out.println(searchInRotatedSortedArray.search(arr, 0));
System.out.println(searchInRotatedSortedArray.search(arr, 5));
System.out.println(searchInRotatedSortedArray.search(arr, 7));
}
}