[抄题]:

You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.

Now, we define a pair (c, d) can follow another pair (a, b) if and only if b < c. Chain of pairs can be formed in this fashion.

Given a set of pairs, find the length longest chain which can be formed. You needn't use up all the given pairs. You can select pairs in any order.

Example 1:

Input: [[1,2], [2,3], [3,4]]
Output: 2
Explanation: The longest chain is [1,2] -> [3,4]

[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]:

基本考虑到了,但是不严谨:首先要确定是坐标型n-1,还是序列型n吧

[英文数据结构或算法,为什么不用别的数据结构或算法]:

lambda:Arrays.sort(pairs, (a,b) -> (a[0] - b[0])); 名称,括号->括号

[一句话思路]:

[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. dp数组不能和题目给的数组搞混

[二刷]:

[三刷]:

[四刷]:

[五刷]:

[五分钟肉眼debug的结果]:

[总结]:

基本考虑到了,但是不严谨:首先要确定是坐标型n-1,还是序列型n吧

[复杂度]:Time complexity: O(n^2) Space complexity: O(n)

[算法思想:迭代/递归/分治/贪心]:

[关键模板化代码]:

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

[代码风格] :

[是否头一次写此类driver funcion的代码] :

[潜台词] :

class Solution {
public int findLongestChain(int[][] pairs) {
//corner case
if (pairs == null || pairs.length == 0) return 0; //initialization: sort, fill the array with 1
Arrays.sort(pairs, (a,b) -> (a[0] - b[0]));
int[] dp = new int[pairs.length];
Arrays.fill(dp, 1); //for loop for i and j
for (int i = 0; i < pairs.length; i++) {
for (int j = i + 1; j < pairs.length; j++) {
dp[j] = Math.max(dp[j], pairs[j][0] > pairs[i][1] ? dp[i] + 1: dp[i]);
}
} return dp[pairs.length - 1];
}
}

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