153. Find Minimum in Rotated Sorted Array

Whatever how we rotate the array, either the left or the right part will be a sorted subarray:

1. If the rightmost element of the left subarray is larger than the rightmost element of the right part, then the pivot point must exist in the right part;
2. Otherwise, the pivot point exists in the left part.

public int findMin(int[] nums) {
    int begin = 0, end = nums.length - 1;
    while( begin < end ) {
        int mid = (begin + end) / 2;
        if( nums[mid] > nums[end] ) // pivot point exists in the right part
            begin = mid + 1;
        else // pivot point exists in the left part
            end = mid;
    }
    return nums[begin];
}