PattySung

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], …, nums[n-1], nums[0], nums[1], …, nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Example 3:

Input: nums = [1], target = 0
Output: -1

<解題>

要在sorted array裡面找target，但數列有被旋轉過，所以我們必須知道哪邊是最小值，才可以去找到target： a. 先確認嘴小值指標left b. 把最小值left設為start，來找到他的區間，如[4,5,6,7,0,1,2]中，0的左側或右側 c. 找target


class Solution {
    public int search(int[] nums, int target) {
        if(nums.length == 0) return -1;

        int left = 0;
        int right = nums.length-1;

        //先確認最小值的指標left
        while(left<right){
            int midpoint = left + (right - left)/2;
            if(nums[midpoint] > nums[right]){
                left = midpoint + 1;
            } else {
                right = midpoint;
            }
        }
        //把最小值left設為start，來找到他的區間，如[4,5,6,7,0,1,2]中，0的左側或右側
        int start = left; //start變成最小值
        left = 0;
        right = nums.length-1;
   
        if(target >= nums[start] && target <= nums[right]){ //target在start和right之間（也就是0的右側）
            left = start;
        } else{
            right = start;
        }

        //找target
        while (left<=right){
            int midpoint = left + (right - left)/2;
            if(nums[midpoint] == target){
                return midpoint;
            } else if(nums[midpoint] > target){
                right = midpoint -1;
            } else {
                left = midpoint +1;
            }

        }
        return -1;
    }
}

Time: O(logN)

Space: O(1)

Leetcode-33. Search in Rotated Sorted Array

<解題>