673.Number of Longest Increasing Subsequence

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.

len 数组记录着以num[i]作为结尾的最长递增序列的长度，cnt 数组记录着以num[i]作为结尾的最长递增序列的个数

 

public int findNumberOfLIS(int[] nums) {
  int n = nums.length, res = 0, max_len = 0;
  int[] len = new int[n];
  int[] cnt = new int[n];
  for(int i = 0; i < n; i++) {
    len[i] = 1;
    cnt[i] = 1;
    for(int j = 0; j < i ; j++) {//对每一元素，都从第一个元素开始与它比较
      if(nums[i] > nums[j]) {
        if(len[i] == len[j] + 1)//比如对7进行遍历时，此时i等于4，j等于3(即遍历到4时)
          cnt[i] += cnt[j];
        if(len[i] < len[j] + 1) {
          len[i] = len[j] + 1;
          cnt[i] = cnt[j];
        }
      }
    }
    if(max_len == len[i])
      res += cnt[i];
    if(max_len < len[i]){
      max_len = len[i];
      res = cnt[i];
    }
  }
  return res;
}