动态规划-673Number of Longest Increasing Subsequence

题目：

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

Example 1:

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

Example 2:

Input: [2,2,2,2,2]Output: 5Explanation: 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.

题意解读：给定一个未排序的整数数组，寻找最长递增子数组的长度

class Solution {    public int findNumberOfLIS(int[] nums) {        //定义两个辅助数组        // len[i]：以nums[i]结尾的,最长递增子序列的长度        // cnt[i]:以nums[i]结尾的，最长递增子序列的条数        //状态转移        //如果nums[i]>nums[j] 并且len[i] = len[j]+1（说明再一次找到最长递增子序列） cnt[i] += cnt[j]        //如果nums[i]>nums[j] 并且len[i] < len[j]+1 (说明有新的最长递增子序列生成) len[i] = len[j]+1;cnt[i] = cnt[j]        int[] len = new int[nums.length];        int[] cnt = new int[nums.length];        int maxLen = 0;int res = 0;        for(int i = 0; i < nums.length; i++){            //初始化len[i],cnt[i]            len[i] = 1;cnt[i] = 1;            for(int j = 0; j < i; j++){                if(nums[i] > nums[j])                    if(len[i] == len[j] + 1)                        cnt[i] += cnt[j];                    else if(len[i] < len[j] + 1){                        len[i] = len[j]+1;//更新最大程度                        cnt[i] = cnt[j];                    }            }            if(maxLen == len[i]) res += cnt[i];//如过len[i]是当前最大长度，将当前cnt[i]累加到结果中去            if(maxLen < len[i]){ //如果len[i]大于当前最大长度，则更新最大长度，并且将结果更新为cnt[i]                maxLen = len[i];                res = cnt[i];            }        }        return res;    }    }