GuilinDev

Lc1304

05 August 2008

1304. Find N Unique Integers Sum up to Zero

给一个n，随便返回一个数组，使其里面有n个不同数字，并加起来为0

方法1. 两边同时找

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class Solution {
    public int[] sumZero(int n) {
        int[] result = new int[n];
        int num = 1;
        for(int i = 0;i < n / 2; i++) {
            result[i] = num;
            result[n - i - 1] = -num;
            num++;
        }
        // 如果中间还剩一个数
        if(n % 2 == 1) {
            result[n / 2] = 0;
        }
        return result;
    }
}

方法2， 找规律

如果

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n = 1, [0]
n = 2, [-1, 1]
n = 3, [-2, 0, 2]
n = 4, [-3, -1, 1, 3]
n = 5, [-4, -2, 0, 2, 4]

可以看出，规律如下：

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A[i] = i * 2 - n + 1

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class Solution {
    public int[] sumZero(int n) {
        int[] A = new int[n];
        for (int i = 0; i < n; ++i)
            A[i] = i * 2 - n + 1;
        return A;
    }
}
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