前言

当要求用户输入数字，用户就是不输入数字，输入字母符号的话，之前写的代码会出现崩溃报错，因为我们没有设计对该种情况的处理方法，今天的课程就教了我们该怎么样处理。

CIN返回false

当int a; cin >> a;，但是输入的是一个字符时，cin会因为输入类型错误返回false值，并使cin失效，可以借助这个机制实现上述功能。

cin.clear()可以恢复cin，再通过cin.get()把缓存区输入的其他的字符取出，就可以进行新一轮输入

习题1

统计鱼的重量，按q中止输入

#include <iostream>
const int Max = 5;
int main()
{
	using namespace std;
	double fish[Max];
	cout << "pls enter the weights of your fish.\n";
	cout << "you may enter up to " << Max << " fish <q to terminate>.\n";
	cout << "fish #1: ";
	int i = 0;
	while (i< Max && cin >> fish[i])
	{
		if (++i < Max)
			cout << "fish #" << i + 1 << ": ";
	}
	double total = 0.0;
	for (int j = 0; j < i; j++)
	{
		total += fish[j];
	}
	if (i == 0)
		cout << "No fish\n";
	else
		cout << total / i << "= average weight of " << i << " fish\n";
	cout << "Done.\n";
}

习题2

统计高尔夫个数，输入错误可以重新输入

#include <iostream>

const int Max = 5;

int main() {
	using namespace std;
	int golf[Max];
	cout << "Pls enter your golf scores.\n";
	cout << "You must enter " << Max << " rounds.\n";
	int i;
	for (i = 0; i < Max; i++)
	{
		cout << "round #" << i + 1 << ": ";
		while (!(cin >> golf[i]))
		{
			cin.clear();
			while (cin.get() != '\n')
				continue;
			cout << "Pls enter a number: ";
		}
	}
	double total = 0.0;
	for (i = 0; i < Max; i++)
		total += golf[i];

	cout << total / Max << " = average score " << Max << "rounds\n";
	return 0;

}

课后习题

可以提前退出的高尔夫

以习题2为基础，加上习题1的功能：即统计高尔夫的分数，如果输错了可以重复输入，但是如果输入的是q则直接中止输入。

例如：

Pls enter your golf scores.
You must enter 5 rounds.
round #1: 20
round #2: w
Pls enter a number: 33
round #3: q
10.6 = average score 5rounds

桂桂答题

#include <iostream>

const int Max = 5;

int main() {
	using namespace std;
	int golf[Max];
	cout << "Pls enter your golf scores.\n";
	cout << "You must enter " << Max << " rounds.\n";
	int i;
	for (i = 0; i < Max; i++)
	{
		cout << "round #" << i + 1 << ": ";
		while (!(cin >> golf[i]))
		{
			cin.clear();
			if (cin.get() == 'q')
				break;
			while (cin.get() != '\n')
				continue;
			cout << "Pls enter a number: ";
		}
	}
	double total = 0.0;
	for (i = 0; i < Max; i++)
		total += golf[i];

	cout << total / Max << " = average score " << Max << "rounds\n";
	return 0;

}

break只能跳出一个循环，在本题中只能够跳出while循环，无法跳出for循环，所以不能退出，令i=Max是为了跳出for循环。

LEETCODE二分查找

课后习题答案

#include <iostream>

const int Max = 5;

int main() {
	using namespace std;
	int golf[Max] = { 0 };  //初始化
	cout << "Pls enter your golf scores.\n";
	cout << "You must enter " << Max << " rounds.\n";
	int i;
	for (i = 0; i < Max; i++)
	{
		cout << "round #" << i + 1 << ": ";
		while (!(cin >> golf[i]))
		{
			cin.clear();
			if (cin.get() == 'q')//判断是否输入为q
			{
				i = Max;
				break;
			}
			while (cin.get() != '\n')
				continue;
			cout << "Pls enter a number: ";
		}
	}
	double total = 0.0;
	for (i = 0; i < Max; i++)
		total += golf[i];

	cout << total / Max << " = average score " << Max << "rounds\n";
	return 0;

}

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size() - 1; // 定义target在左闭右闭的区间里，[left, right]
        while (left <= right) { // 当left==right，区间[left, right]依然有效，所以用 <=
            int middle = left + ((right - left) / 2);// 防止溢出 等同于(left + right)/2
            if (nums[middle] > target) {
                right = middle - 1; // target 在左区间，所以[left, middle - 1]
            } else if (nums[middle] < target) {
                left = middle + 1; // target 在右区间，所以[middle + 1, right]
            } else { // nums[middle] == target
                return middle; // 数组中找到目标值，直接返回下标
            }
        }
        // 未找到目标值
        return -1;
    }
};