【leetcode】Search in Rotated Sorted Array I&II

初始版本

题目描述

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.

解题思路

借用一张图，其实旋转之后就对应了这三种情况

我们能够通过mid与两边的值的比较来得到他们的关系。

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int l = nums.size();
        if (l==0) return -1;
        int left = 0;
        int right = l-1;
        int mid = (left + right) / 2;
        while(left <= right) {
            mid = (left + right) / 2;
            if (nums[mid] == target) return mid;
            else if(nums[mid] < nums[right]){
                //情况1和3
                if (nums[mid] > target || nums[right]<target)
                {
                    right = mid -1;
                }
                else{
                    left = mid +1;
                }
            }
            else{
                if (nums[mid] > target && nums[left]<=target){
                    right = mid -1;
                }
                else{
                    left = mid+1;
                }
            }
        }
        return -1;
    }
};

扩展版本（Search in Rotated Sorted Array II）

题目描述

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).

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

The array may contain duplicates.
即在上面问题的基础上，现在可能出现重复值了

解题思路

就是现在区分1、3和2不光是不等的情况，还有可能相等了，此时只能逐渐增right了（所以最坏情况复杂度也O(n)了）

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int l = nums.size();
        if (l==0) return false;
        int left = 0;
        int right = l-1;
        int mid = (left + right) / 2;
        while(left <= right) {
            mid = (left + right) / 2;
            if (nums[mid] == target) return true;
            else if(nums[mid] < nums[right]){
                //情况1和3
                if (nums[mid] > target || nums[right]<target)
                {
                    right = mid -1;
                }
                else{
                    left = mid +1;
                }
            }
            else if(nums[mid] > nums[right]){
                if (nums[mid] > target && nums[left]<=target){
                    right = mid -1;
                }
                else{
                    left = mid+1;
                }
            }
            else right--;
        }
        return false;
    }
};

References

http://blog.csdn.net/ljiabin/article/details/40453607