C语言刷题 LeetCode 30天挑战 （二）快慢指针法

Write an algorithm to determine if a number is "happy"

A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of thesquares of its digits,

and repeat the process until the number equals 1 (where it will stay, or it loops endlessly in a cycle which does notinclude 1. Those numbers for which this process ends in 1 are happy numbers.

Input:19

Output:true

Explanation

1^2 + 9^2 = 82

8^2 + 2^2 = 68

6^2 + 8^2 = 100

1^2 + 0^2 + 0^2 = 1

循环检测 是不是在1轮回 快慢指针~

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

#define DEBUG

int next_n(int n){
    int r=0;
    while(n != 0){
        int d = n % 10;
        n /= 10;
        r += d*d;
    }
    return r;
}

bool contains(int *history,int size,int n){
    for(int i=0; i<size ;++i){
        if(history[i] == n){
            return true;
        }
    }
    return false;
}

bool ishappy(int n){
    int slow=n;
    int fast=n;
    
    do{
        slow = next_n(slow);
        fast = next_n(next_n(fast));
#ifdef DEBUG
        printf("%d %d \n",slow,fast);
#endif
    } while (slow != fast);

    return false;
}

int main()
{
    ishappy(19);
    if(ishappy)
        printf("yes");
    else printf("no");
    return 0;
} 

slow 和 fast 都初始化为输入的数字 n。

slow 每次移动一步，即计算一次数字的平方和；fast 每次移动两步，计算两次平方和。

在循环中，如果 slow == fast，则表示数字进入了循环，不会到达 1。

但是，在这段代码中，无论循环如何执行，最后的结果总是返回 false，这个逻辑需要进一步调整。