标签：count public target 复杂 二分查找 https value 思路 mina
https://leetcode.com/problems/search-in-rotated-sorted-array/?tab=Description
很好的很经典的题目。今天复习了一下。之前的思路虽然有了，但是对于相等的数字的处理很复杂，容易出错。今天看到了一个很好的解法。
https://discuss.leetcode.com/topic/3538/concise-o-log-n-binary-search-solution
思路是先找出最小的数字的下标，通过 mid > hi，就low = mid+1，否则 hi=mid，来找出。
然后二分查找的时候，用的是偏移过的实际下标。
整体代码：
class Solution { public: int search(int A[], int n, int target) { int lo=0,hi=n-1; // find the index of the smallest value using binary search. // Loop will terminate since mid < hi, and lo or hi will shrink by at least 1. // Proof by contradiction that mid < hi: if mid==hi, then lo==hi and loop would have been terminated. while(lo<hi){ int mid=(lo+hi)/2; if(A[mid]>A[hi]) lo=mid+1; else hi=mid; } // lo==hi is the index of the smallest value and also the number of places rotated. int rot=lo; lo=0;hi=n-1; // The usual binary search and accounting for rotation. while(lo<=hi){ int mid=(lo+hi)/2; int realmid=(mid+rot)%n; if(A[realmid]==target)return realmid; if(A[realmid]<target)lo=mid+1; else hi=mid-1; } return -1; } };
当然了，还有常规的做法，就是这样的。
考虑问题的时候，也不要太过复杂化了。
public int search(int[] A, int target) { int lo = 0; int hi = A.length - 1; while (lo < hi) { int mid = (lo + hi) / 2; if (A[mid] == target) return mid; if (A[lo] <= A[mid]) { if (target >= A[lo] && target < A[mid]) { hi = mid - 1; } else { lo = mid + 1; } } else { if (target > A[mid] && target <= A[hi]) { lo = mid + 1; } else { hi = mid - 1; } } } return A[lo] == target ? lo : -1; }
标签：count public target 复杂 二分查找 https value 思路 mina
原文地址：http://www.cnblogs.com/charlesblc/p/6444027.html
