[LeetCode] Search in Rotated Sorted Array I, II

Search in Rotated Sorted Array I

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.

Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

思路：Search in Rotated Sorted Array I

题目一看就知道是binary search。所以关键点在于每次要能判断出target位于左半还是右半序列。解这题得先在纸上写几个rotated sorted array的例子出来找下规律。Rotated sorted array根据旋转得多少有两种情况：

原数组：0 1 2 4 5 6 7
情况1：  6 7 0 1 2 4 5    起始元素0在中间元素的左边
情况2：  2 4 5 6 7 0 1    起始元素0在中间元素的右边

两种情况都有半边是完全sorted的。根据这半边，当target != A[mid]时，可以分情况判断：

当A[mid] < A[end] < A[start]：情况1，右半序列A[mid+1 : end] sorted
A[mid] < target <= A[end], 右半序列，否则为左半序列。

当A[mid] > A[start] > A[end]：情况2，左半序列A[start : mid-1] sorted
A[start] <= target < A[mid], 左半序列，否则为右半序列

Base case：当start + 1 = end时
假设 2 4:
A[mid] = A[start] = 2 < A[end]，A[mid] < target <= A[end] 右半序列，否则左半序列

假设 4 2:
A[mid] = A[start ] = 4 > A[end],  A[start] <= target < A[mid] 左半序列，否则右半序列

加入base case的情况，最终总结的规律是：

A[mid] =  target, 返回mid，否则

(1) A[mid] < A[end]: A[mid+1 : end] sorted
A[mid] < target <= A[end]  右半，否则左半。

(2) A[mid] > A[end] : A[start : mid-1] sorted
A[start] <= target < A[mid] 左半，否则右半。

递归解法：

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class Solution { public: int search(int A[], int n, int target) { return searchRotatedSortedArray(A, 0, n-1, target); } int searchRotatedSortedArray(int A[], int start, int end, int target) { if(start>end) return -1; int mid = start + (end-start)/2; if(A[mid]==target) return mid; if(A[mid]<A[end]) { // right half sorted if(target>A[mid] && target<=A[end]) return searchRotatedSortedArray(A, mid+1, end, target); else return searchRotatedSortedArray(A, start, mid-1, target); } else { // left half sorted if(target>=A[start] && target<A[mid]) return searchRotatedSortedArray(A, start, mid-1, target); else return searchRotatedSortedArray(A, mid+1, end, target); } } };

递归解法很容易改写成迭代解法：

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class Solution { public: int search(int A[], int n, int target) { int start = 0, end = n-1; while(start<=end) { int mid = start + (end-start)/2; if(A[mid]==target) return mid; if(A[mid]<A[end]) { // right half sorted if(target>A[mid] && target<=A[end]) start = mid+1; else end = mid-1; } else { // left half sorted if(target>=A[start] && target<A[mid]) end = mid-1; else start = mid+1; } } return -1; } };

思路：Search in Rotated Sorted Array II

当有重复数字，会存在A[mid] = A[end]的情况。此时右半序列A[mid-1 : end]可能是sorted，也可能并没有sorted，如下例子。

3 1 2 3 3 3 3
3 3 3 3 1 2 3

所以当A[mid] = A[end] != target时，无法排除一半的序列，而只能排除掉A[end]：

A[mid] = A[end] != target时：搜寻A[start : end-1]

正因为这个变化，在最坏情况下，算法的复杂度退化成了O(n)：

序列 2 2 2 2 2 2 2 中寻找target = 1。

迭代解法：

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class Solution { public: bool search(int A[], int n, int target) { int start = 0, end = n-1; while(start<=end) { int mid = start + (end-start)/2; if(A[mid]==target) return true; if(A[mid]<A[end]) { // right half sorted if(target>A[mid] && target<=A[end]) start = mid+1; else end = mid-1; } else if(A[mid]>A[end]) { // left half sorted if(target>=A[start] && target<A[mid]) end = mid-1; else start = mid+1; } else { end--; } } return false; } };