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，可是在旋转以前是递增有序的，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，因此是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，由于数组是递增的，若是不小于则第一个元素就是最小的元素。若是不小于，初始化high=0,high=len-1,则取中间数mid，若是中间数大于high指向的元素，则说明小数存在后半段，则将high指向mid，不然将high指向mid，继续进行寻找。

只须要判断是否为递增序列或者是否只有一个元素特殊状况便可，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状况，小于判断递增序列。注意：若是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，否则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，可是在旋转以前是递增有序的，找出该数组的的最小值。（数组有重复值）

两道题解法同样，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;
		else
			return 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;
			// 若是三个数字相等须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);
		    }
		}