[抄题]:
Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't one, return 0 instead.
Note:
The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.
Example 1:
Given nums = [1, -1, 5, -2, 3]
, k = 3
,
return 4
. (because the subarray [1, -1, 5, -2]
sums to 3 and is the longest)
Example 2:
Given nums = [-2, -1, 2, 1]
, k = 1
,
return 2
. (because the subarray [-1, 2]
sums to 1 and is the longest)
Follow Up:
Can you do it in O(n) time?
[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]:
以为只有max(max, a)才能取最长,不知道怎么写。其实就是一般思路 sum随着i增加而变长,把i给for一遍就行了。
[一句话思路]:
gap中一下子增加了k时,index也取对应值即可
[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]:
- map.containsKey(sum)没用,必须要sum == k才行
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
必须考虑中间的gap. gap中一下子增加了k时,index也取对应值即可
[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]:
用常量sum才能累加,数组不能:
sum[i] += nums[i]; 相当于每个数都重新加了
[算法思想: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]:
2 sum: hashmap
[代码风格] :
class Solution {
public int maxSubArrayLen(int[] nums, int k) {
//cc
if (nums == null || nums.length == 0) return 0; //ini: hashmap, max
HashMap<Integer, Integer> map = new HashMap<>();
int max = 0, sum = 0; //for loop: 3 conditions
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
//contains, gap, not contains
if (sum == k) max = i + 1;
else if (map.containsKey(sum - k)) max = Math.max(max, i - map.get(sum - k)); if (!map.containsKey(sum)) map.put(sum, i);
} return max;
}
}