624. Maximum Distance in Arrays二重数组中的最大差值距离
2024-09-03 11:45:50
[抄题]:
Given m
arrays, and each array is sorted in ascending order. Now you can pick up two integers from two different arrays (each array picks one) and calculate the distance. We define the distance between two integers a
and b
to be their absolute difference |a-b|
. Your task is to find the maximum distance.
Example 1:
Input:
[[1,2,3],
[4,5],
[1,2,3]]
Output: 4
Explanation:
One way to reach the maximum distance 4 is to pick 1 in the first or third array and pick 5 in the second array.
[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]:
[一句话思路]:
差的绝对值最大有两种情况:最大减最小、最小减最大
[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]:
- 应该理解下:max min都是全组共享的,所以max = Math.max(max, a)很常用
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
[复杂度]:Time complexity: O(n) Space complexity: O(1)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
忘记链表的get方法怎么写了
[关键模板化代码]:
min = Math.min(min, arrays.get(i).get(0));
max = Math.max(max, arrays.get(i).get(arrays.get(i).size() - 1));
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
class Solution {
public int maxDistance(List<List<Integer>> arrays) {
//cc
if (arrays == null || arrays.size() == 0) {
return 0;
} //ini: max min res
int res = Integer.MIN_VALUE;
int min = arrays.get(0).get(0);
int max = arrays.get(0).get(arrays.get(0).size() - 1); //for loop, update max min
for (int i = 1; i < arrays.size(); i++) {
res = Math.max(res, Math.abs(arrays.get(i).get(0) - max));
res = Math.max(res, Math.abs(arrays.get(i).get(arrays.get(i).size() - 1) - min)); min = Math.min(min, arrays.get(i).get(0));
max = Math.max(max, arrays.get(i).get(arrays.get(i).size() - 1));
} //return res
return res;
}
}
[代码风格] :
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