673. Number of Longest Increasing Subsequence

class Solution {
    public int findNumberOfLIS(int[] nums) {
        // 定义 f[i] 为以 nums[i] 结尾的最长递增子序列的长度
        int[] f = new int[nums.length];
        Arrays.fill(f, 1);

        // 定义 g[i] 为以 nums[i] 结尾的最长递增子序列的个数
        int[] g = new int[nums.length];
        Arrays.fill(g, 1);

        int maxLength = 1;
        for (int i = 1; i < nums.length; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[j] < nums[i]) {
                    if (f[j] + 1 > f[i]) {
                        // 能形成一个更长的递增子序列
                        f[i] = f[j] + 1;
                        g[i] = g[j];
                    } else if (f[j] + 1 == f[i]) {
                        g[i] += g[j];
                    }
                }
            }

            maxLength = Math.max(maxLength, f[i]);
        }

        // 注意长度最长的子序列可能以任意元素结尾
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            if (f[i] == maxLength) {
                count += g[i];
            }
        }
        return count;
    }
}

Reference

673. Number of Longest Increasing Subsequence