The following is the Template II code:
int binarySearch(int[] nums, int target){
if(nums == null || nums.length == 0)
return -1;
int left = 0, right = nums.length;
while(left < right){
// Prevent (left + right) overflow
int mid = left + (right - left) / 2;
if(nums[mid] == target){ return mid; }
else if(nums[mid] < target) { left = mid + 1; }
else { right = mid; }
}
// Post-processing:
// End Condition: left == right
if(left != nums.length && nums[left] == target) return left;
return -1;
}I found with right = nums.length - 1 it can pass all test cases for https://leetcode.com/problems/binary-search/, so what is the reason/scenario that need right = nums.length ?