2014
11-18

LeetCode-Search in Rotated Sorted Array[数组]

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.

// LeetCode, Search in Rotated Sorted Array
// 时间复杂度O(log n)，空间复杂度O(1)
class Solution {
public:
int search(int A[], int n, int target) {
int first = 0, last = n;
while (first != last) {
const int mid = (first + last) / 2;
if (A[mid] == target)
return mid;
if (A[first] <= A[mid]) {
if (A[first] <= target && target < A[mid])
last = mid;
else
first = mid + 1;
} else {
if (A[mid] < target && target <= A[last-1])
first = mid + 1;
else
last = mid;
}
}
return -1;
}
};


Java代码:

public class Solution {
public static int search(int[] A, int target) {
return searchBinary(A, 0, A.length-1 , target);
}

public static int searchBinary(int A[],int s,int e,int target){
int mid = (s+e)/2;
if(A[mid] == target)
return mid;
if(s >= e) return -1;
if( A[mid] < A[e] ){ //后半部分是顺序的
if(target > A[mid] && target <= A[e]){
return searchBinary(A, mid+1, e, target);
}else
return searchBinary(A, s, mid-1, target);
}else{ //前半部分顺序的
if(target >= A[s] && target < A[mid])
return searchBinary(A, s, mid-1, target);
else
return searchBinary(A, mid+1, e, target);
}

}
}