如何找到数组中非递减子序列的数量？

Question

给定一个正整数数组，我想找出数组中非递减子序列的个数。

例如，如果数组是{6,7,8,4,5,6}，非递减子序列将是{6},{7},{8},{4},{5},{6},{6,7},{7,8},{4,5},{5,6},{6,7,8},{4,5,6}，所以有12个这样的序列

Answer 1

您可以使用类似于 well-known quadratic solution for the longest increasing subsequence 的动态规划方法。

让 a[i] 成为您的输入数组。设 c[i] 为结束于 a[i] 的非递减子序列的数量。您可以通过查看此类子序列中 a[i] 之前的数字来轻松计算 c[i]。它可以是a[i]之前的任何数字a[j]（即j<i）和不大于（a[j]<=a[i]） .也不要忘记单元素子序列 {a[i]}。这导致以下伪代码：

c[0] = 1
for i = 1..n-1
    c[i] = 1 // the one-element subsequence
    for j = 0..i-1 
        if a[j]<=a[i]
            c[i] += c[j]

另见 Number of all longest increasing subsequences。它只查找最长序列，但我想它也可以修改以计算所有此类序列。

Answer 2

这是一种算法，将列出数字序列中的每个上升子序列：

Set a pointer to the first item, to remember where the rising sequence starts.
Iterate over every item in the array, and for each item:  
    If the current item is not greater than the previous item:  
        Set the pointer to the current item.
    For every n = 1, 2, 3... :
        Save the last n items as a sequence until you reach the pointer.

A 运行-通过您的示例输入 [6,7,8,4,5,6] 将是：

step 1: start=6, current=6, store [6]
step 2: start=6, current=7, comp 7>6=true, store [7], [6,7]
step 3: start=6, current=8, comp 8>7=true, store [8], [7,8], [6,7,8]
step 4: start=6, current=4, comp 4>8=false, set start to current item, store [4]
step 5: start=4, current=5, comp 5>4=true, store [5], [4,5]
step 6: start=4, current=6, comp 6>5=true, store [6], [5,6], [4,5,6]

result: [6], [7], [6,7], [8], [7,8], [6,7,8], [4], [5], [4,5], [6], [5,6], [4,5,6]

例如javascript：（注意：slice()函数用于创建数组的硬拷贝）

function rising(array) {
    var sequences = [], start = 0;
    for (var current = 0; current < array.length; current++) {
        var seq = [], from = current;
        if (array[current] < array[current - 1]) start = current;
        while (from >= start) {
            seq.unshift(array[from--]);
            sequences.push(seq.slice());
        }
    }
    return sequences;
}

var a = rising([6,7,8,4,5,6]);
document.write(JSON.stringify(a));

如果您希望结果按照您在问题中所写的顺序排列：[6],[7],[8],[4],[5],[6],[6,7],[7,8],[4,5],[5,6],[4,5,6],[6,7,8] 然后将 sequences 设为二维数组并将每个序列 seq 存储在 sequences[seq.length] 中.

如何找到数组中非递减子序列的数量？

How can I find the number of non-decreasing subsequences in an array?

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subsequence