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✨ topic

673. Number of longest increasing subsequences


🤣 digression

The number of the longest increasing subsequence, not the length of the longest increasing subsequence.No, no, no (￣▽￣)


Don’t ask me how I know (manual funny)

300. Eldest son sequence

class Solution {

    public int lengthOfLIS(int[] nums) {

        int[] dp = new int[nums.length];

        for(int i = 0; i < nums.length; i++) { dp[i] =1;
        }

        int res = 1;
        for(int i = 1; i < nums.length; i++) {for(int j = 0; j < i; j++) {if(nums[j] < nums[i]) {
                    dp[i] = Math.max(dp[i],dp[j] + 1);
                }
            }
            res = Math.max(res,dp[i]);
        }

        returnres; }}Copy the code


😉 serious point, solution idea

The answer to this question isThe number of longest increasing subsequences, so we have to know what the longest is, to find the longest sequence number is!


Using dynamic programming, setint[] dp = new int[nums.length]; Array record lengthTo set upint[] counts = new int[nums.length]; Record the number of


So how does the state transition ❔

If there is familiarity300. Eldest son sequenceAs your friends may know, I don’t talk nonsense

Because we want to increment, so ⏬ whenDp [j] + 1 > dp[I] + 1 > dp[I] If true, dp[j] should not be preceded by dp[I]. If dp[I] is longer than dp[I], then I need to update the length

But if theDp [j] + 1 == dp[I], dp[j] + 1 == dp[I]


🌈 code implementation

class Solution {
    public int findNumberOfLIS(int[] nums) {
        int len = nums.length;
        if (len <= 1) return len;
        int[] dp = new int[len]; 
        int[] counts = new int[len]; 

        // The length must be at least 1
        Arrays.fill(counts, 1);

        int maxLen = 0;

        for (int i = 0; i < len; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[j] < nums[i]) {
                    if (dp[j] + 1 > dp[i]) {
                        dp[i] = dp[j] + 1;
                        counts[i] = counts[j];
                    }else if (dp[j] + 1 == dp[i]) {
                        counts[i] += counts[j];
                    }
                }
            }
            maxLen = Math.max(maxLen,dp[i]);
        }

        // Select maxLen from maxLen
        int res = 0;
        for (int i = 0; i < len; ++i) {
            if(dp[i] == maxLen) { res += counts[i]; }}returnres; }}Copy the code


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