//基础二分
int binary_search(int[] nums, int target) {
    int left = 0, right = nums.length - 1; 
    while(left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] < target) {
            left = mid + 1;
        } else if (nums[mid] > target) {
            right = mid - 1; 
        } else if(nums[mid] == target) {
            // 直接返回
            return mid;
        }
    }
    // 直接返回
    return -1;
}
//寻找左边界
int left_bound(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] < target) {
            left = mid + 1;
        } else if (nums[mid] > target) {
            right = mid - 1;
        } else if (nums[mid] == target) {
            // 别返回，锁定左侧边界
            right = mid - 1;
        }
    }
    // 判断 target 是否存在于 nums 中
    if (left < 0 || left >= nums.length) {
        return -1;
    }
    // 判断一下 nums[left] 是不是 target
    return nums[left] == target ? left : -1;
}
//寻找右边界
int right_bound(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] < target) {
            left = mid + 1;
        } else if (nums[mid] > target) {
            right = mid - 1;
        } else if (nums[mid] == target) {
            // 别返回，锁定右侧边界
            left = mid + 1;
        }
    }
    // 判断 target 是否存在于 nums 中
    // if (left - 1 < 0 || left - 1 >= nums.length) {
    //     return -1;
    // }
    
    // 由于 while 的结束条件是 right == left - 1，且现在在求右边界
    // 所以用 right 替代 left - 1 更好记
    if (right < 0 || right >= nums.length) {
        return -1;
    }
    return nums[right] == target ? right : -1;
}