Leetcode 153. Find Minimum in rotated Sorted Array

Suppose a sorted array is rotated on 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.

Assume no duplicate exists in the array.

Analysis:

This problem gives an ordered array to find the smallest element in the array, in fact, to find the first position less than the last element appears,

For example 4 5 6 0 1 2 given in the title,

The last element is 2,

Just find the location where the first element less than 2 appears.

The first element is not used as target, because there is a limit to flipping an array, which is to completely not flip, that is 0 1 4 5 6 7,

In this case, an error occurs if the first element is used as the target.

Find the first element that <= Targrt, and the last element is the target.

 Public classSolution {//find the first element that <= Targrt, and the last element is the target.     Public intFindmin (int[] nums) {        if(Nums.length = = 0){            return0; }        intStart = 0, end =nums.length; inttarget = Nums[nums.length-1]; intmid;  while(Start + 1 <end) {Mid= start + (End-start)/2; if(Nums[mid] <=target) {End=mid; }Else{Start=mid; }        }        if(Nums[start] <=target) {            returnNums[start]; }Else{            returnNums[end]; }    }}

Leetcode 153. Find Minimum in rotated Sorted Array