673. Number of Longest Increasing Subsequence

问题描述：

Given an unsorted array of integers, find the number of longest increasing subsequence.

Example 1:

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

 

Example 2:

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.

 

Note: Length of the given array will be not exceed 2000 and the answer is guaranteed to be fit in 32-bit signed int.

 

解题思路：

可以用一个数组lens存储到这个数字位置最长的自增序列的长度，用一个数组cnt存储以这个数字为结尾的最长的递增序列的个数。

更新lens比较好更新：比较nums[i] 和 nums[j]的大小。

更新cnt时需注意，需要比较lens[i] 和 lens[j]的大小：

　　注意此时前提是nums[j] < nums[i]

　　1. 若lens[i] < lens[j] + 1 也就是说 lens[i] 记录的不为最长的，至少比lens[j] , nums[i]的序列要短

　　此时我们更新lens[i] = lens[j] +1, cnt[i] = cnt[j]

　　2.若lens[i] == lens[j] + 1，说明当前这个序列又是满足条件的最长子序列 所以 cnt[i] += cnt[j]

 

代码：

class Solution {
public:
    int findNumberOfLIS(vector<int>& nums) {
        vector<int> lens(nums.size(), 1);
        vector<int> cnt(nums.size(), 1);
        int maxLen = 0;
        for(int i = 0; i < nums.size(); i++){
            for(int j = 0; j < i; j++){
                if(nums[i] > nums[j]){
                    if(lens[i] == lens[j] + 1)
                        cnt[i] += cnt[j];
                    else if(lens[i] < lens[j] + 1){
                        lens[i] = lens[j]+1;
                        cnt[i] = cnt[j];
                    }
                }
            }
            maxLen = max(maxLen, lens[i]);
        }
        int ret = 0;
        for(int i = 0; i <lens.size(); i++){
            if(lens[i] == maxLen)
                ret += cnt[i];
        }
        return ret;
    }
};