153.154. Find Minimum in Rotated Sorted Array

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.

分析：二分法，首先将left和right进行比较，然后将left和mid相比较，确定递增序列，然后更改left或right的值。

public class Solution {   public int findMin(int[] nums) {    if(nums==null)    return 0;    int n=nums.length;    int left=0,right=n-1;    int mid;    while(left<right){    mid=left+(right-left)/2;    if(nums[left]<=nums[right])    return nums[left];    else if(nums[left]>nums[right]&&nums[mid]<nums[right]){    right=mid;    }else if(nums[left]>nums[right]&&nums[left]<=nums[mid]){    left=mid+1;    }    }    return nums[left];    }}

154. Find Minimum in Rotated Sorted Array II

数组中可能存在重复的数字，此时有一种情况，无法判断左边和右边哪一边是递增的，即中间的等于左边的和右边的时候，此时只能将left+1，right-1.

public class Solution {    public int findMin(int[] nums) {        if(nums==null)            return 0;        int n=nums.length;        int left=0,right=n-1;        int mid;        while(left<right){            mid=left+(right-left)/2;            if(nums[left]<nums[right])                return nums[left];            else if(nums[left]==nums[right]&&nums[left]==nums[mid]){                left++;                right--;            }            else if(nums[left]>=nums[right]&&nums[mid]>=nums[left])                left=mid+1;            else if(nums[left]>=nums[right]&&nums[mid]<=nums[right])                right=mid;                    }        return nums[left];    }}