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.
解题思路:正常人都能想到一个解法,遍历所有值,然后返回。这个做法虽然也能accepted,但是违背了出题者的本意。通过一番思考和上网查阅后,学到一个算法,感谢ljiabin大神。他提供的一个求解图让我豁然开朗。通过这个图,当只考虑升序的时候,基本只有这三种情况,我们只要把所有的情景都能够利用二分法处理,那么这道题很容易。三种情况的区分主要是mid(中点)处在不同位置的结果,正常的二分法则按照图一来处理,其余两种的不同便是mid的不同。通过left right mid 三个中点的不同值来区分这三种情况。在处理返回值的时候,这里有一个简易的做法,递归的结束都在函数的开始的时候处理,这样的做法减少了代码的冗余。
代码如下:
public int search(int[] nums, int target) { int len = nums.length-1; return binarySearch(nums, 0, len, target); } public int binarySearch(int[] nums, int left, int right, int target){ int mid = (left+right)/2; if(left>right)return -1; if(nums[left]==target)return left; if(nums[mid]==target)return mid; if(nums[right]==target)return right; if(nums[left]<nums[right]){ if(target<nums[left]||target>nums[right]){ return -1; }else if(target>nums[mid]){ return binarySearch(nums, mid+1, right-1, target); }else{ return binarySearch(nums, left+1, mid-1, target); } }else if(nums[left]<nums[mid]){ if(target>nums[left]&&target<nums[mid]){ return binarySearch(nums, left+1, mid-1, target); }else{ return binarySearch(nums, mid+1, right-1, target); } }else{ if(target<nums[right]&&target>nums[mid]){ return binarySearch(nums, mid+1, right-1, target); }else{ return binarySearch(nums, left+1, mid-1, target); } } }
http://blog.csdn.net/ljiabin/article/details/40453607
该文章中还提出了一个更简洁的算法:
public class Solution { public int search(int[] A, int target) { int l = 0; int r = A.length - 1; while (l <= r) { int mid = (l + r) / 2; if (target == A[mid]) return mid; if (A[l] <= A[r]) { if (target < A[mid]) r = mid - 1; else l = mid + 1; } else if (A[l] <= A[mid]) { if (target > A[mid] || target < A[l]) l = mid + 1; else r = mid - 1; } else { if (target < A[mid] || target > A[r]) r = mid - 1; else l = mid + 1; } } return -1; } }