【leetcode刷题笔记】Search in Rotated Sorted Array

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.

题解：还是按照二分的方法找target。

如果A[l] < A[mid]，说明mid以左有序且都小于mid，如下图所示：这种情况下如果target在l和mid之间，那么需要把r重新置为mid；其他情况都需要到mid右端继续搜索。



2.如果A[l] >= A[mid], 说明mid以右有序且都大于mid，如下图所示，如果target在mid和r之间，那么需要把l重新置为mid；其他情况都需要到mid左端继续搜索。



当l + 1 = r的时候，只要检查l和r所指向的元素是否等于target即可。

代码如下：

public class Solution {
public int search(int[] A, int target) {
int l = 0;
int r = A.length - 1;

while(l + 1< r){
int mid = l + (r-l)/2;
if(A[mid] == target)
return mid;
if(A[l]< A[mid] ){
if(A[mid] >= target && A[l] <= target)
r = mid;
else {
l = mid;
}
}
else {
if(target >= A[mid] && target <= A[r])
l = mid;
else {
r = mid;
}
}
}

if(target == A[l])
return l;
if(target == A[r])
return r;
return -1;
}
}