Thursday, June 9, 2016

equal sum – Moonstone


equal sum – Moonstone
http://www.1point3acres.com/bbs/thread-191823-1-1.html
2 一个数组里找某个index, 使sum[:i] == sum[i+1:], 也是经典题。我一开始用了O(n) space, follow up就是优化成了O(1). 
这里代码写的有点慢,但都在没给提示的前提下bug free了。
3 上道题的变种,此时要求数组和带有权重,每个nums[i]需要乘以一个weight, 这个weight等于和某个index的距离。
eg:
nums = [1, 3, 5, 7, 8]
假如当前处理到nums[2], 则leftsum = 1 * 2 + 3 * 1 = 5, rightsum = 7 * 1 + 8 * 2 = 23

public int EqualSum(int[] input)
{
    if ((input == null) || (input.Length <= 1))
    {
        return -1;
    }
 
    int sum = 0;
    int len = input.Length;
 
    for (int i = 0; i < len; i++)
    {
        sum += input[i];
    }
 
    if (sum % 2 == 1)
    {
        return -1;
    }
 
    int halfSum = sum / 2;
    int s = 0;
 
    int j = 0;
    while (j < len)
    {
        s += input[j];
        if (s == halfSum)
        {
            return j;
        }
        j++;
    }
 
    return j;
 
}
 
public int EqualSumWithWeigh(int[] input)
{
    if ((input == null) || (input.Length <= 1))
    {
        return -1;
    }
 
    int sum = 0;
    int len = input.Length;
    int[] p = new int[len];
 
    for (int i = 0; i < len; i++)
    {
        sum += input[i];
        p[i] = sum;
    }
 
    int k = 0;
    int leftSum = 0;
    int rightSum = 0;
 
    int j = k + 1;
    while (j < len)
    {
        rightSum += (j - k) * input[j];
        j++;
    }
 
    while (k <= len)
    {
        if (leftSum == rightSum)
        {
            return k;
        }
 
        k++;
        //leftSum - 0..k-1, k, k+1..N-1
        leftSum += p[k - 1];
        rightSum -= (p[len - 1] - p[k - 1]);
    }
 
    return k;
}
 
public void TestEqualSum()
{
    int[] input1 = new int[] { -2, 3, 2, 1, 2 };
    int index1 = EqualSum(input1);
}
 
public void TestEqualSumWithWeigh()
{
    int[] input1 = new int[] { 1, 3, 5, 7, 8 };
    int r1 = EqualSumWithWeigh(input1);
}
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