for 循环变量处理的基本理解

• j ++ 和 ++ j 什么区别

对于for (int j = 0; j < 10; j ++ ) 和 for (int j = 0; j < 10; ++ j ), j 的作用是相同的：就是把j的值加上1；对于赋值情况，z = ++ j 和 z = j ++, 假设j = 0: 前者z = 1, 后者z = 0

• 如果变量j在for循环内部改变的话，会怎样?

for (int j = 0; j < 10; j ++ )
    j = 4;
// 这种情况会出现死循环， TLE： 因为j ++ = 5 永远小于10， 会一直循环;
// int j = 0 指挥运行一次，之后就看for循环内部的j的变量，如果满足j < 10， 那么执行 j ++; 当j >= 10 时跳出循环;
// 所以并不是 一直是 j 从 0 ~ 9，要看for里面变量的改变和前面j++的部分结合;

• 例子

#include <iostream>
#include <algorithm>

using namespace std;

int n;

int main()
{
    for (int j = 0; j < 10; j ++ )
    {
        cout << "初始j:" << j << " ";

        int k  = j + 2;
        cout << "k:" << k << " ";

        int q = j ++;
        cout << "q:" << q << " "; 
        // 考虑初始情况  j = 0
        // 当前q = j，q = 0，但是j = j + 1, 这里j = 1

        int z = ++ j;
        cout << "z:" << z << " ";
        //因为j已经是1， 这里z = j + 1, 所以z = 2

        j =  k - 1;
        cout << "末尾j:" << j << " ";
        cout << endl;
    }
    return 0;
}

• 结果

初始j:0 k:2 q:0 z:2 末尾j:1 
初始j:2 k:4 q:2 z:4 末尾j:3 
初始j:4 k:6 q:4 z:6 末尾j:5 
初始j:6 k:8 q:6 z:8 末尾j:7 
初始j:8 k:10 q:8 z:10 末尾j:9

• 例题

/*
 * @lc app=leetcode id=38 lang=cpp
 *
 * [38] Count and Say
 *
 * https://leetcode.com/problems/count-and-say/description/
 *
 * algorithms
 * Easy (41.40%)
 * Likes:    888
 * Dislikes: 6930
 * Total Accepted:    312.9K
 * Total Submissions: 750.8K
 * Testcase Example:  '1'
 *
 * The count-and-say sequence is the sequence of integers with the first five
 * terms as following:
 * 
 * 
 * 1.     1
 * 2.     11
 * 3.     21
 * 4.     1211
 * 5.     111221
 * 
 * 
 * 1 is read off as "one 1" or 11.
 * 11 is read off as "two 1s" or 21.
 * 21 is read off as "one 2, then one 1" or 1211.
 * 
 * Given an integer n where 1 ≤ n ≤ 30, generate the n^th term of the
 * count-and-say sequence.
 * 
 * Note: Each term of the sequence of integers will be represented as a
 * string.
 * 
 * 
 * 
 * Example 1:
 * 
 * 
 * Input: 1
 * Output: "1"
 * 
 * 
 * Example 2:
 * 
 * 
 * Input: 4
 * Output: "1211"
 * 
 */
class Solution {
public:
//从前往后模拟做
    string countAndSay(int n) {
        string s = "1"; // n = 1 时
        for (int i = 2; i <= n; i ++ ) // 从 n = 2 到 n
        {
            string new_s;
            for (int j = 0; j < s.size(); j ++ )
            {
                int k = j; //从j开始寻找，判断后面有多少个相同字符
                while (k < s.size() && s[k] == s[j]) k ++ ; //求出当前j对应的k值(k为相同字符的后一位标号)
                new_s += to_string(k - j) + s[j];
                //问题：这里不是从j = 0, 1, 2 ... 一直遍历吗， 那么new_s += 会出现多加的的情况
                j = k - 1;
                //cout << j << " ";
            }
            s = new_s;
        }
        return s;      
    }
};