Find Min in Rotated Array-优快云博客

在未知旋转点的已排序数组中找到最小元素。给定数组[4,5,6,7,0,1,2]，返回0。解决方法包括暴力搜索和二分查找，其中二分查找能更有效地排除一半的数。" 126994472,11922784,OpenCV实战：面部特征提取与滤镜应用,"['计算机视觉', '人工智能', 'python', 'opencv', '图像处理']

Find Min in Rotated Array

在旋转数组中找到最小值。

Suppose a sorted array 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).

Find the minimum element.

You may assume no duplicate exists in the array.

Example

Given [4,5,6,7,0,1,2] return 0

Solution:

Brutal:

    public int findMin(int[] num) {
        if (num == null || num.length == 0) {
            return 0;
        }
        int min = num[0];
        for (int i : num) {
            if (i < min) {
                min = i;
                break;
            }
        }
        return min;
    }

二分查找：

    public int findMin(int[] num) {
        if (num == null || num.length == 0) {
            return 0;
        }
        int start = 0;
        int end = num.length - 1;
        int mid;
        int min = num[start];
        while (start + 1 < end) {
            mid = start + ((end - start) >> 1);
            if (num[mid] < min) {
                min = num[mid];
                end = mid;
            } else {
                start = mid;
            }
        }
        return min < num[end] ? min : num[end];
    }

    public int findMin(int[] num) {
        if (num == null || num.length == 0) {
            return 0;
        }
        int start = 0;
        int end = num.length - 1;
        int mid;
        while (start + 1 < end) {
            mid = start + ((end - start) >> 1);
            if (num[end] < num[mid]) {
                start = mid;
            } else {
                end = mid;
            }
        }
        return num[start] < num[end] ? num[start] : num[end];
    }

思路：

暴力：两段都单调递增，而且第一段比第二段要高。所以转点就是最小值。

二分查找：每次排除一半的数。

防止没有分段的情况，如{1,2,3}，从end判断。谨慎使用mid + 1和mid - 1.