LintCode 62. Search in Rotated Sorted Array 原创Java参考解答
问题描述
http://www.lintcode.com/en/problem/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.
Example
For [4, 5, 1, 2, 3] and target=1, return 2.
For [4, 5, 1, 2, 3] and target=0, return -1.
解题思路
题目是在一种已按大小排序但被旋转过的数组（Rotated Sorted Array，RST）中查找目标值target。
通过题目给出的例子，可以看到这种数组的特点是由小变大，到了某一个位置可能会突然下降，然后再次逐渐由小变大。值得注意的是数组如[1, 2, 3] 其实也算作Rotated Sorted Array，它假定只是把第一个元素和自己做位置调换。
这道题运用二分法解决。分两次二分法，第一次找到这个RST中的最小元素位置，从而通过比较target和最小元素大小关系，判断target可能是在的数组中左右哪段区间。接着，再一次二分法查找该区间是否有目标值。
参考代码
public class Solution {
/**
*@param A : an integer rotated sorted array
*@param target : an integer to be searched
*return : an integer
*/
public int search(int[] A, int target) {
if (A == null || A.length == 0) {
return -1;
}
int minPosition = 0;
int start = 0;
int end = A.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (A[mid] > A[end]) {
start = mid;
} else {
end = mid;
}
}
if (A[start] < A[end]) {
minPosition = start;
} else {
minPosition = end;
}
if (A[minPosition] == target) {
return minPosition;
} else {
if (A[A.length - 1] < target) {
start = 0;
end = minPosition - 1;
} else {
start = minPosition;
end = A.length - 1;
}
}
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (A[mid] == target) {
return mid;
} else if (A[mid] < target) {
start = mid;
} else {
end = mid;
}
}
if (A[start] == target) {
return start;
}
if (A[end] == target) {
return end;
}
return -1;
}
}