leetcode154. Find Minimum in Rotated Sorted Array II

154. Find Minimum in Rotated Sorted Array II

Suppose an array sorted in ascending order 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.

The array may contain duplicates.

解法一

二分查找，如果遇到mid和end相等，则end-1。

public class Solution {    public int findMin(int[] nums) {        if (nums == null || nums.length == 0) {            return -1;        }        int start = 0;        int end = nums.length - 1;        int mid;        while (start + 1 < end) {            mid = start + (end - start) / 2;            if (nums[mid] == nums[end]) {                end--;            } else if (nums[mid] < nums[end]) {                end = mid;            } else {                start = mid;            }        }        return nums[start] <= nums[end] ? nums[start] : nums[end];    }}

解法二

一个for循环，找最小值，因为最坏的情况是[1,1,1,……,1]中有一个0，时间复杂度为O(n)，相当于一个for循环

public class Solution {    public int findMin(int[] nums) {        if (nums == null || nums.length == 0) {            return -1;        }        int min = nums[0];        for (int i = 1; i < nums.length; i++) {            if (nums[i] < min) {                min = nums[i];            }        }        return min;    }}