leetcode解题之153&154. Find Minimum in Rotated Sorted Array版（在旋转的数组中查找最小数字）

153.Find Minimum in Rotated Sorted Array

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.

You may assume no duplicate exists in the array.

给定一个数组，数组可能被旋转了，但是在旋转之前是递增有序的，找出该数组的的最小值。（数组没有重复值）

复杂度等价于一个二分查找，是O(logn)，空间上只有四个变量维护二分和结果，所以是O(1)。

采用二分查找的方法进行查找。数组的第一个元素要大于最后一个元素，因为数组是递增的，如果不小于则第一个元素就是最小的元素。如果不小于，初始化high=0,high=len-1,则取中间数mid，如果中间数大于high指向的元素，则说明小数存在后半段，则将high指向mid，否则将high指向mid，继续进行寻找。

只需要判断是否为递增序列或者是否只有一个元素特殊情况即可，添加一句，if (nums[low] <= nums[high]) return nums[low];继续二分查找。等号判断只有一个元素情况，小于判断递增序列。注意：如果中间元素小于最左边元素，则右部分为有序数组。说明此时中间元素已经在最小值右侧，不然不会小于最左侧元素

154. Find Minimum in Rotated Sorted Array II

Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

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.

给定一个数组，数组可能被旋转了，但是在旋转之前是递增有序的，找出该数组的的最小值。（数组有重复值）

两道题解法一样，有重复值只要判断 low high mid三者指向的元素是否相等，相等就退化为O（N）复杂度。

public int search(int[] nums, int target) {if (nums == null || nums.length < target)return 0;int low = 0;int res = 0;int high = nums.length - 1;if (nums[low] <= nums[high])res = Arrays.binarySearch(nums, target);else {int m = findMin(nums);System.out.println(m);res = Arrays.binarySearch(nums, 0, m, target);if (res < 0)res = Arrays.binarySearch(nums, m, high + 1, target);}if (res < 0)return -1;elsereturn res;}public int findMin(int[] nums) {// ask interviewer which value should return:// Integer.MIN_VALUE or throw a Exception.if (nums == null || nums.length == 0)return -1;int low = 0;int mid = 0;int high = nums.length - 1;while (nums[low] >= nums[high]) {// 表示找到if (high - low == 1) {mid = high;break;}mid = (low + high) / 2;// 如果三个数字相等需要遍历整个数字查找最小值if (nums[mid] == nums[low] && nums[mid] == nums[high])return MinOrder(nums, low, high);// 处于递增子序列中，最小值在右侧if (nums[mid] >= nums[low])low = mid;else// 处于递减子序列中，最小值在左侧high = mid;}return nums[mid];}// 遍历整个数组求最小值private int MinOrder(int[] nums, int low, int high) {int min = nums[low];for (int i = low + 1; i <= high; i++)min = Math.min(min, nums[i]);return min;}

网上参考：http://www.programcreek.com/2014/03/leetcode-find-minimum-in-rotated-sorted-array-ii-java/

public class Solution {   public int findMin(int[] num) {    return findMin(num, 0, num.length-1);} public int findMin(int[] num, int left, int right){    if(right==left){        return num[left];    }    if(right == left +1){        return Math.min(num[left], num[right]);    }    // 3 3 1 3 3 3     int middle = (right-left)/2 + left;    // already sorted    if(num[right] > num[left]){        return num[left];    //right shift one    }else if(num[right] == num[left]){        return findMin(num, left+1, right);    //go right        }else if(num[middle] >= num[left]){        return findMin(num, middle, right);    //go left        }else{        return findMin(num, left, middle);    }}