剑指offer 11. 旋转数组的最小数字

题目描述

把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个递增排序的数组的一个旋转，输出旋转数组的最小元素。例如，数组 [3,4,5,1,2] 为 [1,2,3,4,5] 的一个旋转，该数组的最小值为 1。

示例 1：

输入：[3,4,5,1,2]
输出：1

示例 2：

输入：[2,2,2,0,1]
输出：0

解法

Python3

class Solution:
    def minArray(self, numbers: List[int]) -> int:
        l, r = 0, len(numbers) - 1
        while l < r:
            m = (l + r) >> 1
            if numbers[m] > numbers[r]:
                l = m + 1
            elif numbers[m] < numbers[r]:
                r = m
            else:
                r -= 1
        return numbers[l]

Java

class Solution {
    public int minArray(int[] numbers) {
        int l = 0, r = numbers.length - 1;
        while (l < r) {
            int m = (l + r) >>> 1;
            if (numbers[m] > numbers[r]) {
                l = m + 1;
            } else if (numbers[m] < numbers[r]) {
                r = m;
            } else {
                --r;
            }
        }
        return numbers[l];
    }
}

JavaScript

/**
 * @param {number[]} numbers
 * @return {number}
 */
var minArray = function (numbers) {
  let l = 0,
    r = numbers.length - 1;
  while (l < r) {
    let m = (l + r) >>> 1;
    if (numbers[m] > numbers[r]) {
      l = m + 1;
    } else if (numbers[m] < numbers[r]) {
      r = m;
    } else {
      --r;
    }
  }
  return numbers[l];
};

C++

class Solution {
public:
    int minArray(vector<int>& numbers) {
        int left = 0, right = numbers.size() - 1;
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (numbers[mid] > numbers[right]) {
                left = mid;
            } else if (numbers[mid] < numbers[right]) {
                right = mid;
            } else {
                --right;
            }
        }
        return min(numbers[left], numbers[right]);
    }
};