【算法分析与设计】【第十一周】673. Number of Longest Increasing Subsequence

题目来源：673：https://leetcode.com/problems/number-of-longest-increasing-subsequence/description/

动态规划基础训练。

673. Number of Longest Increasing Subsequence

题目大意

求出一个无序整数数组的最长递增子序列的数量。

题目有点绕，其实分解一下就是：
（1）求出最长递增子序列的长度len
（2）求长度为len的递增子序列有几条
题目就是求这个条数。

例
Input: [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequence are [1, 3, 4, 7] and [1, 3, 5, 7].

Input: [2,2,2,2,2]
Output: 5
Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences’ length is 1, so output 5.

思路

其实本质是最长递增子序列问题（LIS）。
只要在求最长递增子序列长度的同时计数就可以了。

1. 求LIS的长度
   （1）维护数组length，length[i]用于记录以第i个元素为最后一个元素的递增子序列的长度；
   （2）考虑第j个元素：
   将第0~j-1个元素中有小于第j个元素的元素放入集合S，在集合S中找出length最大的si，将第j个元素放到si后面。

2. 求最长递增子序列长度的同时计数

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LIS简单回顾

关于LIS可参考《算法概论》P157

算法

for j = 1, 2, ..., n  L(j) = 1 + max{L(i):(i, j)∈E}return maxjL(j)

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解题代码

class Solution {public:  int findNumberOfLIS(vector<int>& nums) {    int n = nums.size();    // handle special case    if (n <= 1) return n;    int length[n] = {1};  // length[i] : the length of the seq ends with the ith element    int count[n] = {1};  // count[i] : the number of seqs of length i    for (int i = 0; i < n; i++) length[i] = count[i] = 1;    for (int j = 0; j < n; j++) {      for (int i = 0; i < j; i++) {        if (nums[j] > nums[i]) {          if (length[i] >= length[j]) {            // add nums[j] into the seq which ends with the ith element, and the length becomes length[i]+1            length[j] = length[i] + 1;            count[j] = count[i];          } else if (length[i] + 1 == length[j]) {            // merge two seqs            count[j] += count[i];          }        }      }    }    int maxLen = 0;    for (int i = 0; i < n; i++) {      if (length[i] > maxLen) maxLen = length[i];    }    int re = 0;    for (int i = 0; i < n; i++) {      if (length[i] == maxLen) {        re += count[i];      }    }    return re;  }};

时间复杂度

O(n^2)

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