153. Find Minimum in Rotated Sorted Array

Leetcode

https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/

題目

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

• [4,5,6,7,0,1,2] if it was rotated 4 times.

• [0,1,2,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Example 1:

Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 

解答

• 方法一

如果不限制用 O(log N) 解的話，只要遍歷矩陣找出下一個元素大於目前元素時，就是矩陣旋轉的地方

var findMin = function (nums) {
    for (let i = 0; i < nums.length; i++) {
        if (nums[i + 1] < nums[i]) {
            return nums[i + 1];
        }
    }

    return nums[0];
};

空間複雜度： O(n)

• 方法二

由於矩陣在旋轉點前、後都是由小到大排序，所以最快可以用二分搜尋法達到 O(lon N) 的時間複雜度

var findMin = function (nums) {
    let low = 0;
    let high = nums.length - 1;

    while (low <= high) {
        const mid = Math.floor((low + high) / 2);
        // 找到旋轉點
        if (nums[mid + 1] < nums[mid]) {
            return nums[mid + 1];
        } else if (nums[mid] >= nums[0]) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }

    return nums[0];
};