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33. 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.
问题： 给定一个已排序的数组，该数组以某一个元素作为支点做了旋转，在改旋转后数组中搜索值。
已排序数组的搜索，自然会想到二分搜索。将旋转到后面的部分用负数表示下标，便可以正常使用二分搜索。
需要注意的是对比元素值 和 目标值时，记得将 下标转会正数 ( i + len ) % len 。
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81. 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.
问题： 若 Search in Rotated Sorted Array 中的数组存在重复元素，如何解决原问题？
重复元素对于上面算法唯一一个影响，就是算法实现时候，判断是否有旋转需要更严谨一点。
第一题代码：
1 int search(vector<int>& nums, int target) { 2 3 int len = (int)nums.size(); 4 5 if (len == 0){ 6 return -1; 7 } 8 9 if ( len == 1 ){ 10 return (nums[0] == target) ? 0 : -1; 11 } 12 13 int realL; 14 int realR; 15 if (nums[0] > nums[len-1]) { 16 for ( int i = 0 ; i < len ; i++){ 17 if (nums[i] > nums[i+1]){ 18 realR = i; 19 break; 20 } 21 } 22 23 realL = realR - len + 1; 24 }else{ 25 realL = 0; 26 realR = len - 1; 27 } 28 29 while( realL < realR ){ 30 31 if (realL + 1 == realR){ 32 int idxL = ( realL + len ) % len; 33 int idxR = ( realR + len ) % len; 34 35 if (nums[idxL] == target){ 36 return idxL; 37 } 38 39 if (nums[idxR] == target){ 40 return idxR; 41 } 42 43 return -1; 44 } 45 46 int mid = ( realL + realR ) / 2; 47 int idx = ( mid + len ) % len; 48 49 if (nums[idx] == target){ 50 return idx; 51 } 52 53 if (nums[idx] < target){ 54 realL = mid; 55 }else{ 56 realR = mid; 57 } 58 } 59 60 // Actually, program will never step to here. It will return value previously. 61 return -1; 62 }
第二题代码：
bool search(vector<int>& nums, int target) { int len = (int)nums.size(); if (len == 0){ return false; } if ( len == 1 ){ return (nums[0] == target) ? 1 : 0; } int realL; int realR; int ii = 0; while (ii < len - 1) { if (nums[ii] <= nums[ii + 1]) { ii++; continue; }else{ break; } } if (ii == len - 1 ) { realL = 0; realR = len - 1; }else{ realR = ii; realL = realR - len + 1; } while( realL < realR ){ if (realL + 1 == realR){ int idxL = ( realL + len ) % len; int idxR = ( realR + len ) % len; if (nums[idxL] == target){ return true; } if (nums[idxR] == target){ return true; } return false; } int mid = ( realL + realR ) / 2; int idx = ( mid + len ) % len; if (nums[idx] == target){ return true; } if (nums[idx] < target){ realL = mid; }else{ realR = mid; } } // Actually, program will never step to here. It will return value previously. return -1; }
[LeetCode] Search in Rotated Sorted Array I (33) && II (81) 解题思路
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原文地址：http://www.cnblogs.com/TonyYPZhang/p/5079159.html
