673. Number of Longest Increasing Subsequence

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本文介绍了解决LeetCode第673题——求最长递增子序列个数的方法。通过动态规划的方式，实现了一个Java程序来计算给定整数数组中最长递增子序列的个数。

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673. Number of Longest Increasing Subsequence

题目描述：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.

题目大意：给定一个未排序的int数组，找出最长子序列的个数

代码

package DP;

/**
* @Author OovEver
* @Date 2017/12/15 10:06
*/
public class LeetCode673 {
  public int findNumberOfLIS(int[] nums) {
      int n = nums.length;
      int res = 0;
      int max_len = 0;
//        以nums[i]结尾的最长子序列的长度
      int[] len = new int[n];
//        以nums[i]结尾的最长子序列的个数
      int[] count = new int[n];
      for(int i=0;i<n;i++) {
          len[i] = count[i] = 1;
          for(int j=0;j<i;j++) {
              if (nums[i] > nums[j]) {
                  if(len[i]==len[j]+1) count[i] += count[j];
                  if (len[i] < len[j] + 1) {
                      len[i] = len[j] + 1;
                      count[i] = count[j];
                  }
              }
          }
          if(max_len==len[i]) res += count[i];
          if (max_len < len[i]) {
              max_len = len[i];
              res = count[i];
          }
      }
      return res;
  }
}