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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public class Solution { /** * modified the binary search.<br> *when the array is rotated, we first find the minimum and then do binary search in appropriate subarray.<br> * @param A --Integer array, search in this array * @param target --int, searched number * * @return int. If the target is found, return the index, otherwise return -1. * @author Averill Zheng * @version 2014-06-05 * @since JDK 1.7 */ public
int search(int[] A, int
target) { int
length = A.length, index = -1; if(length > 0){ int
tempIndex = 0; if(A[0] <= A[length - 1]) //normal order tempIndex = Arrays.binarySearch(A, target); //index = (tempIndex >= 0) ? tempIndex : -1; else{ //find the min first int
min = findMin(A, 0, length - 1); if(target >= A[0]) tempIndex = Arrays.binarySearch(A, 0, min, target); else tempIndex = Arrays.binarySearch(A, min, length, target); } index = (tempIndex < 0) ? -1
: tempIndex; } return
index; } public
int findMin(int[] A, int
i, int j){ int
min = i, left = i, right = j; if(A[i] > A[j]){ while(left < right - 1){ int
mid = left + (right - left) / 2; if(A[left] < A[mid]) left = mid; else right = mid; } min = (A[left] < A[right])? left : right; } return
min; }} |
leetcode online judge is also accepted the code if we use the linear search method to find the minimum as follows:
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public int findMin(int[] A, int
i, int j){ int
left = i, right = j; while(left < right){ if(A[left] > A[right]) ++left; else --right; } return
left;} |
leetcode--Search in Rotated Sorted Array,布布扣,bubuko.com
leetcode--Search in Rotated Sorted Array
原文地址：http://www.cnblogs.com/averillzheng/p/3774364.html
