153. Find Minimum in rotated Sorted Array

First, the topic

1, examining

　　

2. Analysis

After an ordered integer array is flipped into a two-segment increment ordinal group, the smallest element is found.

Second, the answer

1, Ideas:

Method One,

The minimum element is obtained by the dichotomy method.

①, Nums[start] < Nums[end], the array is ordered; return Nums[start];

②, Nums[start] < nums[mid], the first half is ordered, then start = mid + 1;

Otherwise the second half is ordered, then end = mid;

③, finally, return to Nums[start];

     Public intFindmin (int[] nums) {             intLen =nums.length; if(len = = 1)            returnNums[0]; intStart = 0, end = len-1;  while(Start <end) {                        //already orderly            if(Nums[start] <Nums[end]) {                returnNums[start]; }//int mid = (start + end)/2; //May be overflow            intMID = start + ((End-start) >> 1); //first half ordered            if(Nums[start] <Nums[mid]) Start= Mid + 1; //The first half is not required;            ElseEnd=mid; }               returnNums[start]); }    

Method Two,

Directly using linear method to find;

     Public int findmin (int[] nums) {                for (int i = 1; i < nums.length; i++) { c9/>if(Nums[i] < nums[i-1])                return  nums[i];        }                 return nums[0];    }

153. Find Minimum in rotated Sorted Array