二分查找对应三种区间的写法:

// 二分查找 --- [left, right]
    // 数组已经是有序的了!
    public static int binarySerach1(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int left = 0, right = nums.length-1;
        while (left <= right) {
            // 防止溢出 等同于(left + right)/2
            int mid = left + (right-left)/2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] > target) {
                // target 在左区间，所以[left, middle - 1]
                right = mid-1;
            } else {
                // target 在右区间，所以[middle + 1, right]
                left = mid+1;
            }
        }

        return -1;
    }

    // 二分查找 --- [left, right)
    // 数组已经是有序的了!
    int binarySearch2(int[] nums, int target){
        if(nums == null || nums.length == 0)
            return -1;
        // 定义target在左闭右开的区间里，即：[left, right)
        int left = 0, right = nums.length;
        // 因为left == right的时候，在[left, right)是无效的空间，所以使用 <
        while(left < right){
            int mid = left + (right - left) / 2;
            if(nums[mid] == target){
                return mid;
            }
            else if(nums[mid] < target) {
                //  target 在右区间，在[middle + 1, right)中
                left = mid + 1;
            }
            else {
                // target 在左区间，在[left, middle)中
                right = mid;
            }
        }

        // Post-processing:
        // End Condition: left == right
        if(left != nums.length && nums[left] == target) return left;
        return -1;
    }

    // 二分查找 --- (left, right)
    // 数组已经是有序的了!
    int binarySearch3(int[] nums, int target) {
        if (nums == null || nums.length == 0)
            return -1;

        int left = 0, right = nums.length - 1;
        while (left + 1 < right){
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                //  target 在右区间，在(middle, right)中
                left = mid;
            } else {
                // target 在左区间，在(left, middle)中
                right = mid;
            }
        }

        // Post-processing:
        // End Condition: left + 1 == right
        if(nums[left] == target) return left;
        if(nums[right] == target) return right;
        return -1;
    }