• 홀수만 처리 가능!

int size = 5;

int[][] arr = new int[size][size];
		
int middle = size/2;

int i=0, j=middle;
for(int num=1 ; num<=size*size ; num++) {
	arr[i][j] = num;
	
	//행 감소
	i--;
	if(i < 0)
		i = size-1;
	
	//열 증가
	j = (++j)%size;
	//아래 표현과 같다.
	/*j++;
	if(j >= size)
		j = 0;*/

		//배수이면 행은 1 증가, 열은 그대로
	if(num%size == 0) {
		i = (i+2)%size;
		j--;
		if(j < 0)
			j = size-1;
	}
}

참조

참조

• 일반적인 구현

	public static int binarySearch(int[] arr, int target) {
		int left = 0;
		int right = arr.length - 1;
		int mid = 0;
		
		while(left <= right) {
			mid = (left + right) / 2;
			
			if(arr[mid] < target)
				left = mid + 1;
			else if (arr[mid] > target)
				right = mid - 1;
			else
				return mid;
		}
		
		return Integer.MIN_VALUE;
	}

• 재귀적 구현

	public static int binarySearchRecursive(int[] arr, int target, int left, int right) {
		if(left > right)
			return Integer.MIN_VALUE;
		
		int mid = (left + right) / 2;
		if(arr[mid] < target)
			return binarySearchRecursive(arr, target, left + 1, right);
		else if (arr[mid] > target)
			return binarySearchRecursive(arr, target, left, right - 1);
		else
			return mid;
	}

• A ＜ X ≤ B 조건을 검색. (반환값이 index+1 이므로)

	public static int binarySearch(ArrayList<Integer> list, int target) {

		int left = 0;
		int right = list.size() - 1;
		int middle = 0;
		
		while(left <= right) {
			
			middle = (left + right) / 2;
			
			if(target < list.get(middle)) {
				right = middle - 1;
			}
			else {
				left = middle + 1;
			}
			

			if(list.get(middle) <= target && middle < list.size() - 1) {
				if(target < list.get(middle + 1)) {
					return list.get(list.size() > middle + 1 ? middle + 1 : list.size() - 1);
				}
			}
		}
		
		return -1;
	}