[LeetCode] 673. Number of Longest Increasing Subsequence 最长递增序列的个数

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.

class Solution {
public:
int findNumberOfLIS(vector<int>& nums) {
int res = 0, mx = 0, n = nums.size();
vector<int> len(n, 1), cnt(n, 1);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (nums[i] <= nums[j]) continue;
if (len[i] == len[j] + 1) cnt[i] += cnt[j];
else if (len[i] < len[j] + 1) {
len[i] = len[j] + 1;
cnt[i] = cnt[j];
}
}
if (mx == len[i]) res += cnt[i];
else if (mx < len[i]) {
mx = len[i];
res = cnt[i];
}
}
return res;
}
};

class Solution {
public:
int findNumberOfLIS(vector<int>& nums) {
int res = 0, mx = 0, n = nums.size();
vector<int> len(n, 1), cnt(n, 1);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (nums[i] <= nums[j]) continue;
if (len[i] == len[j] + 1) cnt[i] += cnt[j];
else if (len[i] < len[j] + 1) {
len[i] = len[j] + 1;
cnt[i] = cnt[j];
}
}
mx = max(mx, len[i]);
}
for (int i = 0; i < n; ++i) {
if (mx == len[i]) res += cnt[i];
}
return res;
}
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/673

Longest Increasing Subsequence

Longest Continuous Increasing Subsequence

https://leetcode.com/problems/number-of-longest-increasing-subsequence/

https://leetcode.com/problems/number-of-longest-increasing-subsequence/discuss/107318/C%2B%2B-DP-with-explanation-O(n2)

https://leetcode.com/problems/number-of-longest-increasing-subsequence/discuss/107293/JavaC%2B%2B-Simple-dp-solution-with-explanation

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posted @ 2017-09-27 21:06  Grandyang  阅读(9170)  评论(2编辑  收藏  举报