[LeetCode][Java] Search in Rotated Sorted Array
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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.
题意:
假设对于一个有序数组,实现对其利用某种先验旋转,而且你是还不知道
(i.e., 0 1 2 4 5 6 7
might become 4 5 6 7 0 1 2
).
给定一个目标值,在这个数组中取搜寻该目标值,如果找到就返回这个目标值的标号,否则就返回-1.
可以假设该数组中没有重复元素.
算法分析:
参考http://pisxw.com/algorithm/Search-in-Rotated-Sorted-Array.html
该题依旧是二分查找的变形,具体来说,假设数组是A,每次左边缘为l,右边缘为r,还有中间位置是m。在每次迭代中,分三种情况:
(1)如果target==A[m],那么m就是我们要的结果,直接返回;
(2)如果A[m]<A[r],那么说明从m到r一定是有序的(没有受到rotate的影响),那么我们只需要判断target是不是在m到r之间,如果是则把左边缘移到m+1,否则就target在另一半,即把右边缘移到m-1。
(3)如果A[m]>=A[r],那么说明从l到m一定是有序的,同样只需要判断target是否在这个范围内,相应的移动边缘即可
AC代码:
<span style="font-family:SimSun;font-size:12px;">public class Solution { public int search(int[] A, int target) { if(A==null||A.length==0) return -1; int l=0,r=A.length-1; while(l<=r) { int mid=(l+r)/2; if(target==A[mid]) return mid; if(A[mid]<A[r]) { if(target>A[mid]&&target<=A[r]) l=mid+1; else r=mid-1; } else { if(target>=A[l]&&target<A[mid]) r=mid-1; else l=mid+1; } } return -1; }}</span>
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