[抄题]:

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.

[暴力解法]:

时间分析:

空间分析:

[优化后]:

时间分析:

空间分析:

[奇葩输出条件]:

[奇葩corner case]:

[思维问题]:

不知道为什么len[i] == len[j] + 1:因为可以间隔相加。

也不知道为什么是DP:原来小人是间隔着跳的。

[一句话思路]:

长度一个数组、数量一个数组,两个分开算

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

  1. 如果出现了新的最长数组,count需要和最大长度一起换

[二刷]:

[三刷]:

[四刷]:

[五刷]:

[五分钟肉眼debug的结果]:

[总结]:

count length分开算

[复杂度]:Time complexity: O(n) Space complexity: O(n)

[英文数据结构或算法,为什么不用别的数据结构或算法]:

[算法思想:递归/分治/贪心]:贪心

[关键模板化代码]:

count更新或相加:

if (nums[j] < nums[i]) {
if (length[j] + 1 > length[i]) {
length[i] = length[j] + 1;
//renew cnt[i]
count[i] = count[j];
}else if (length[j] + 1 == length[i]) {
count[i] += count[j];
}
}
}

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

LIS本身

[代码风格] :

class Solution {
public int findNumberOfLIS(int[] nums) {
//cc
if (nums == null || nums.length == 0) return 0; //ini: length[], count[], res
int n = nums.length, res = 0, max_len = 0;
int[] length = new int[n];
int[] count = new int[n]; //for loop: i, nums[j] < nums[i], count j, max_length
for (int i = 0; i < n; i++) {
//; not ,
length[i] = 1; count[i] = 1;
for (int j = 0; j < i; j++) {
if (nums[j] < nums[i]) {
if (length[j] + 1 > length[i]) {
length[i] = length[j] + 1;
//renew cnt[i]
count[i] = count[j];
}else if (length[j] + 1 == length[i]) {
count[i] += count[j];
}
}
}
if (length[i] > max_len) {
max_len = length[i];
//renew cnt[i]
res = count[i];
}
else if (length[i] == max_len) res += count[i];
} return res;
}
}

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