剑指offer:面试题8、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。输入一个非递减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。
NOTE：给出的所有元素都大于0，若数组大小为0，请返回0。

代码示例

public class Offer8 {
    public static void main(String[] args) {
        int[] nums = {3,4,5,1,2};
        int[] nums2 = {1,1,1,1,0,1,1};
        Offer8 testObj = new Offer8();
        System.out.println(testObj.minNumInRotateArray(nums));
        System.out.println(testObj.minNumInRotateArray(nums2));
        System.out.println(testObj.minNumInRotateArray2(nums2));
    }

    //数组中没有重复数字的情况
    public int minNumInRotateArray(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }
        int low = 0;
        int high = nums.length - 1;
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (nums[mid] <= nums[high]) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        return nums[low];
    }

    //数组中有重复数字的情况
    public int minNumInRotateArray2(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }
        int low = 0;
        int high = nums.length - 1;
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (nums[low] == nums[mid] && nums[mid] == nums[high]) {
                return minNumber(nums, low, high);
            } else if (nums[mid] <= nums[high]) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        return nums[low];
    }

    private int minNumber(int[] nums, int low, int high) {
        for (int i = low; i < high; i++) {
            if (nums[i] > nums[i + 1]) {
                return nums[i + 1];
            }
        }
        return nums[low];
    }
}