673. Number of Longest Increasing Subsequence

Description:

• Difficulty:Medium
• Total Accepted:7.1K
• Total Submissions:19.3K
• Contributor: fishercoder

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.

Note:Length of the given array will be not exceed 2000 and the answer is guaranteed to be fit in 32-bit signed int.

Thought:

 len[k]: 表示以 nums[k] 为末尾的最长子序列长度。
 max[k]: 表示以  nums[k] 为末尾的最长子序列个数

对于每个num[i]，遍历数组中他前面的所有num，找到比他小的num[j]，如果此时num[j]长度大于当前num[i]的长度，则num[i]可以排在num[j]后面，即len[i]=len[j]+1。如果此时num[i]的长度与num[j] + 1的长度相等，说明这是以num[i]结尾的另一个最长子串，子序列有max[i] += max[j]。

Code:

class Solution {
public:
	int findNumberOfLIS(vector<int>& nums) {
		int length = nums.size();
		int l = 0;
		int count = 0;
		vector<int> len(length + 1, 1);
		vector<int> max(length + 1, 1);
		for (int i = 1; i < length; i++) {
			for (int j = 0; j<i; j++) {
				if (nums[j] < nums[i]) {
					if (len[j] + 1 > len[i]) {
						len[i] = len[j] + 1;
						max[i] = max[j];
					}
					else if (len[j] + 1 == len[i]) {
						max[i] += max[j];
					}
				}
			}
		}
		for (int i = 0; i < length; i++) l = l > len[i] ? l : len[i];
		for (int i = 0; i < length; i++) {
			if (len[i] == l) count += max[i];
		}
		return count;
	}
};