[leetcode]-633. Sum of Square Numbers(C语言）

Given a non-negative integer c, your task is to decide whether there're two integers a and b such that a² + b² = c.

Example 1:

Input: 5
Output: True
Explanation: 1 * 1 + 2 * 2 = 5

Example 2:

Input: 3
Output: False

Approach #1 Brute Force [Time Limit Exceeded]

bool judgeSquareSum(int c) {
    int i,j;
    for(i=0;i*i<=c;i++)
        for(j=0;j*j<=c;j++)
        {
            if(i*i+j*j==c)
                return true;
        }
    return false;
}

此方法虽然简单，但是给定一个比较大的数时，往往在指定时间内求不出结果。

Approach #2 Better Brute Force [Time Limit Exceeded]

We can improve the last solution, if we make the following observation. For any particular $a$ chosen, the value of $b$ required to satisfy the equation a2+b2=ca^2 + b^2 = ca​2​​+b​2​​=c will be such that b2=c−a2b^2 = c - a^2b​2​​=c−a​2​​. Thus, we need to traverse over the range (0,c)(0, \sqrt{c})(0,√​c​​​) only for considering the various values of $a$. For every current value of $a$ chosen, we can determine the corresponding b2b^2b​2​​ value and check if it is a perfect square or not. If it happens to be a perfect square, $c$ is a sum of squares of two integers, otherwise not.

Now, to determine, if the number c−a2c - a^2c−a​2​​ is a perfect square or not, we can make use of the following theorem: "The square of nthn^{th}n​th​​ positive integer can be represented as a sum of first $n$ odd positive integers." Or in mathematical terms:

n2=1+3+5+...+(2∗n−1)=∑1n(2∗i−1)n^2 = 1 + 3 + 5 + ... + (2*n-1) = \sum_{1}^{n} (2*i - 1)n​2​​=1+3+5+...+(2∗n−1)=∑​1​n​​(2∗i−1).

To look at the proof of this statement, look at the L.H.S. of the above statement.

$1+3+5+...+(2\ast n-1)=$

$(2\ast 1-1)+(2\ast 2-1)+(2\ast 3-1)+...+(2\ast n-1)=$

$2\ast (1+2+3+....+n)-(1+1+...ntimes)=$

$2\ast n\ast (n+1)/2-n=$

$n\ast (n+1)-n=$

n2+n−n=n2n^2 + n - n = n^2n​2​​+n−n=n​2​​

根据以上原理，第一次写出的代码总是运行错误，后来才发现一行代码的位置放错导致逻辑出错。

bool judgeSquareSum(int c) { 
   int i,b;
//    int j=1,sum=0;    该行代码应该放在循环内部，否则对每一个i，j和sum的值将一直增加，
    for(i=0;i*i<=c;i++)
    {
        b=c-i*i;
        int j=1,sum=0;
        while(sum<b)
        {
            sum=sum+j;
            j+=2;
        }
        if(sum==b)
            return true;
    }
    return false;
}

但还是运行时间过长。

Approach #3 Using sqrt function[Time Limit Exceeded]

Algorithm

Instead of finding if c−a2c - a^2c−a​2​​ is a perfect square using sum of odd numbers, as done in the last approach, we can make use of the inbuilt $sqrt$ function and check if c−a2\sqrt{c - a^2}√​c−a​2​​​​​ turns out to be an integer. If it happens for any value of $a$ in the range [0,c][0, \sqrt{c}][0,√​c​​​], we can return a True value immediately.

bool judgeSquareSum(int c) {
    int i;
    double b;
    for(i=0;i*i<=c;i++)
    {
        b=sqrt(c-i*i);
        if(b==(int)b)
            return true;
    }
    return false;
}

还是时间太长。

最后的代码如下：

bool judgeSquareSum(int c) {
    int left=0,right=(int)sqrt(c);
    while(left<=right)
    {
        int res=left*left+right*right;
        if(res<c)
            left++;
        else if(res>c)
            right--;
        else
            return true;
    }
    return false;
}