0673. Number of Longest Increasing Subsequence (M)

题目

Given an integer array nums, return the number of longest increasing subsequences.

Example 1:

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

Example 2:

Input: nums = [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.

Constraints:

• 0 <= nums.length <= 2000
• -10^6 <= nums[i] <= 10^6

题意

统计给定数组中最长递增子序列的个数。

思路

动态规划。建两个数组len和cnt：len[i]表示以第i个整数为结尾的递增子序列的长度，cnt[i]表示以第i个整数为结尾的递增子序列的个数。目标就是求len[i]最大时的cnt[i]的值。

代码实现

Java

class Solution {
    public int findNumberOfLIS(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }
        int[] len = new int[nums.length];
        int[] cnt = new int[nums.length];
        len[0] = 1;
        cnt[0] = 1;
        for (int i = 1; i < nums.length; i++) {
            int maxLen = 0, maxCnt = 1;
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    if (len[j] > maxLen) {
                        maxLen = len[j];
                        maxCnt = cnt[j];
                    } else if (len[j] == maxLen) {
                        maxCnt += cnt[j];
                    }
                }
            }
            len[i] = maxLen + 1;
            cnt[i] = maxCnt;
        }
        int maxLen = 0, maxCnt = 0;
        for (int i = 0; i < nums.length; i++) {
            if (len[i] > maxLen) {
                maxLen = len[i];
                maxCnt = cnt[i];
            } else if (len[i] == maxLen) {
                maxCnt += cnt[i];
            }
        }
        return maxCnt;
    }
}