标签：tor text img 中间 ret exists ram hit lap
62. Search in Rotated Sorted Array【medium】
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.
For [4, 5, 1, 2, 3] and target=1, return 2.
For [4, 5, 1, 2, 3] and target=0, return -1.
O(logN) time
错误解法：
1 class Solution { 2 public: 3 /* 4 * @param A: an integer rotated sorted array 5 * @param target: an integer to be searched 6 * @return: an integer 7 */ 8 int search(vector<int> A, int target) { 9 if (A.size() == 0) { 10 return -1; 11 } 12 13 int start = 0; 14 int end = A.size() - 1; 15 16 while (start + 1 < end) { 17 int mid = start + (end - start) / 2; 18 19 if (A[mid] == target) { 20 return mid; 21 } 22 else if (A[mid] < target) { 23 // 5 6 1 2 3 4 24 if (A[mid] > A[start]) { 25 start = mid; 26 } 27 // 5 6 1 2 3 4 28 else { // mid < target && mid <= start 29 end = mid; 30 } 31 } 32 else if (A[mid] > target) { 33 34 if (A[mid] < A[end]) { // mid > target && mid < end 35 end = mid; 36 } 37 // 3 4 5 6 1 2 找3 38 else { // mid > target && mid >= end 39 start = mid; 40 } 41 } 42 } 43 44 if (A[start] == target) { 45 return start; 46 } 47 48 if (A[end] == target) { 49 return end; 50 } 51 52 return -1; 53 } 54 };
一开始思考的方向就不对，搞晕了……
解法一：
1 class Solution { 2 public: 3 /* 4 * @param A: an integer rotated sorted array 5 * @param target: an integer to be searched 6 * @return: an integer 7 */ 8 int search(vector<int> A, int target) { 9 if (A.size() == 0) { 10 return -1; 11 } 12 13 int start = 0; 14 int end = A.size() - 1; 15 16 while (start + 1 < end) { 17 int mid = start + (end - start) / 2; 18 19 if (A[mid] == target) { 20 return mid; 21 } 22 23 if (A[mid] >= A[start]) { 24 if (A[start] <= target && target <= A[mid]) { 25 end = mid; 26 } 27 else { 28 start = mid; 29 } 30 } 31 else { 32 if (A[mid] <= target && target <= A[end]) { 33 start = mid; 34 } 35 else { 36 end = mid; 37 } 38 } 39 } 40 41 if (A[start] == target) { 42 return start; 43 } 44 45 if (A[end] == target) { 46 return end; 47 } 48 49 return -1; 50 } 51 };
借鉴网上的一个图，可以清楚的归纳一下思路，那么代码就好写了。
这个图参考了：http://fisherlei.blogspot.com/2013/01/leetcode-search-in-rotated-sorted-array.html
解法二：
1 class Solution { 2 public: 3 int search(int A[], int n, int target) { 4 return searchRotatedSortedArray(A, 0, n-1, target); 5 } 6 7 int searchRotatedSortedArray(int A[], int start, int end, int target) { 8 if(start>end) return -1; 9 int mid = start + (end-start)/2; 10 if(A[mid]==target) return mid; 11 12 if(A[mid]<A[end]) { // right half sorted 13 if(target>A[mid] && target<=A[end]) 14 return searchRotatedSortedArray(A, mid+1, end, target); 15 else 16 return searchRotatedSortedArray(A, start, mid-1, target); 17 } 18 else { // left half sorted 19 if(target>=A[start] && target<A[mid]) 20 return searchRotatedSortedArray(A, start, mid-1, target); 21 else 22 return searchRotatedSortedArray(A, mid+1, end, target); 23 } 24 } 25 };
解法三：
1 class Solution { 2 public: 3 int search(int A[], int n, int target) { 4 int start = 0, end = n-1; 5 while(start<=end) { 6 int mid = start + (end-start)/2; 7 if(A[mid]==target) return mid; 8 9 if(A[mid]<A[end]) { // right half sorted 10 if(target>A[mid] && target<=A[end]) 11 start = mid+1; 12 else 13 end = mid-1; 14 } 15 else { // left half sorted 16 if(target>=A[start] && target<A[mid]) 17 end = mid-1; 18 else 19 start = mid+1; 20 } 21 } 22 return -1; 23 } 24 };
解法二和解法三参考了：http://bangbingsyb.blogspot.com/2014/11/leetcode-search-in-rotated-sorted-array.html
思路如下：
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
最后总结出：
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。
62. Search in Rotated Sorted Array【medium】
标签：tor text img 中间 ret exists ram hit lap
原文地址：http://www.cnblogs.com/abc-begin/p/7552275.html
