159. Find Minimum in Rotated Sorted Array 【medium】

159. Find Minimum in Rotated Sorted Array 【medium】

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.

Notice

You may assume no duplicate exists in the array.

Have you met this question in a real interview?

Yes

Example

Given [4, 5, 6, 7, 0, 1, 2] return 0

解法一：

 class Solution {
 public:
     /*
      * @param nums: a rotated sorted array
      * @return: the minimum number in the array
      */
     int findMin(vector<int> nums) {
         int start = ;
         int end = nums.size() - ;

         while (start +  < end) {
             int mid = start + (end - start) / ;

             //1 2 3 4 5 6
             if (nums[mid] > nums[start] && nums[mid] < nums[end]) {
                 end = mid;
             }
             // 3 4 5 6 1 2
             else if (nums[mid] > nums[start] && nums[mid] > nums[end]) {
                 start = mid;
             }
             // 5 6 1 2 3 4
             else if (nums[mid] < nums[start] && nums[mid] < nums[end]) {
                 //start = mid;
                 end = mid;
             }
         }

         return nums[start] > nums[end] ? nums[end] : nums[start];
     }
 };

注意分数组的奇偶去考查。比如[1, 2, 3, 4, 5]和[1, 2, 3, 4, 5, 6]是不同的。

解法二：

 public class Solution {
     /**
      * @param nums: a rotated sorted array
      * @return: the minimum number in the array
      */
     public int findMin(int[] nums) {
         if (nums == null || nums.length == ) {
             return -;
         }

         int start = , end = nums.length - ;
         int target = nums[nums.length - ];

         // find the first element <= target
         while (start +  < end) {
             int mid = start + (end - start) / ;
             if (nums[mid] <= target) {
                 end = mid;
             } else {
                 start = mid;
             }
         }
         if (nums[start] <= target) {
             return nums[start];
         } else {
             return nums[end];
         }
     }
 }

大神解法就是给力！