646. Maximum Length of Pair Chain 最长的链条长度
[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]:
- 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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