leetcode--33. Search in Rotated Sorted Array

Suppose an array sorted in ascending order 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.

我的方法：

与discussion中第二种一致。

step1：确定pivot在哪一侧

step2：根据pivot所在边的特点，进行mid的确定

这里格外注意边界的问题：

if(nums[mid] < nums[tail]) { //pivot is on the left

if(tar > nums[tail] || tar < nums[mid]) tail = mid;

都是需要注意的点。否则会引起死循环。

class Solution {public:    int search(vector<int>& nums, int tar) {        int len = nums.size();        if(len == 0) return -1;        int head = 0;        int tail = len-1;        while(head < tail){            int mid = head + (tail-head)/2;            if(nums[mid] == tar) return mid;            if(nums[mid] < nums[tail]) { //pivot is on the left                if(tar > nums[tail] || tar < nums[mid]) tail = mid;                else head = mid+1;            }            else{                if(tar < nums[head] || tar > nums[mid]) head = mid+1;                else tail = mid;            }        }        return nums[tail] == tar? tail : -1;    }};

方案二：

discuss中的第一种方法：

step1：用一个while循环，找到pivot的坐标

step2：还原原始下标，二分法查找

int realmid=(mid+rot)%n;

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;    }};