最长的正面子阵列

Question

数组A []仅包含'1'和'-1'

构造数组B,其中B [i]是从j开始并在i结束的最长连续子阵列的长度,其中 j < i and A[j] + .. + A[i] > 0

明显的O(n ^ 2)解决方案是:

for (int i = 0; i < A.size(); ++i) {
    j = i-1;
    sum = A[i];
    B[i] = -1; //index which fills criteria not found
    while ( j >=0 ) {
        sum += A[j];
        if (sum > 0)
            B[i] = i - j + 1;
        --j;
    }
}

我正在寻找O(n)解决方案.

Answer 1

诀窍是要认识到我们只需要找到满足 的最小 j (A[0] + ... + A[j-1]) == (A[0] + ... + A[i]) - 1。 A[j] + ... + A[i]与 相同(A[0] + ... + A[i]) - (A[0] + ... + A[j-1])，所以一旦我们找到合适的 j，j 和 i 之间的和将为 1。任何较早的 j 都不会产生正值，任何较晚的 j 都不会给我们最长的可能序列。如果我们跟踪第一次达到每个连续负值的位置，那么我们可以轻松地为任何给定的 i 查找正确的 j 。

这是一个 C++ 实现：

vector<int> solve(const vector<int> &A)
{
    int n = A.size();
    int sum = 0;
    int min = 0;
    vector<int> low_points;
    low_points.push_back(-1);
    // low_points[0] is the position where we first reached a sum of 0
    // which is just before the first index.
    vector<int> B(n,-1);
    for (int i=0; i!=n; ++i) {
        sum += A[i];
        if (sum<min) {
            min = sum;
            low_points.push_back(i);
            // low_points[-sum] will be the index where the sum was first
            // reached.
        }
        else if (sum>min) {
            // Go back to where the sum was one less than what it is now,
            // or go all the way back to the beginning if the sum is
            // positive.
            int index = sum<1 ? -(sum-1) : 0;
            int length = i-low_points[index];
            if (length>1) {
                B[i] = length;
            }
        }
    }
    return B;
}