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.
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?
给一个数组nums和一个目标值k，找出子数组和是k的最大长度，如果没有返回0. 要求O(n)时间复杂度。
解法1：双指针，双层循环计算所有的组合，判断是否和为k，如果是，更新max_len。时间复杂度高，TLE
解法：循环数组，用一个变量 cur_sum 记录到目前为止所有数组的和，如果等于k则更新max_len，在用一个 map 记录累加和的index，技巧：因为是求最长数组，所以一个和只记录第一次的index，以后出现的位置靠后，就不记录了。如果cur_sum在hashmap 中，表示当前位置去掉hashmap中记录的cur_sum - k的 index 的和等于k, 用两个index的差更新max_len。
Java:
public int maxSubArrayLen(int[] nums, int k) {
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int max = 0;
int sum=0;
for(int i=0; i<nums.length; i++){
sum += nums[i];
if(sum==k){
max = Math.max(max, i+1);
}
int diff = sum-k;
if(map.containsKey(diff)){
max = Math.max(max, i-map.get(diff));
}
if(!map.containsKey(sum)){
map.put(sum, i);
}
}
return max;
}
Python: Time: O(n)， Space: O(n)
class Solution(object):
def maxSubArrayLen(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
sums = {}
cur_sum, max_len = 0, 0
for i in xrange(len(nums)):
cur_sum += nums[i]
if cur_sum == k:
max_len = i + 1
elif cur_sum - k in sums:
max_len = max(max_len, i - sums[cur_sum - k])
if cur_sum not in sums:
sums[cur_sum] = i # Only keep the smallest index.
return max_len
Python: wo
class Solution():
def maxSubarry(self, nums, k):
m = {0: -1}
sm = 0
for i in range(len(nums)):
sm += nums[i]
if sm not in m:
m[sm] = i
if sm - k in m:
max_len = max(max_len, i - m[sm-k])
return max_len
C++:
class Solution {
public:
int maxSubArrayLen(vector<int>& nums, int k) {
int sum = 0, res = 0;
unordered_map<int, int> m;
for (int i = 0; i < nums.size(); ++i) {
sum += nums[i];
if (sum == k) res = i + 1;
else if (m.count(sum - k)) res = max(res, i - m[sum - k]);
if (!m.count(sum)) m[sum] = i;
}
return res;
}
};
类似题目：
[LeetCode] 53. Maximum Subarray 最大子数组
[LeetCode] 209. Minimum Size Subarray Sum 最短子数组之和
[LeetCode] 560. Subarray Sum Equals K 子数组和为K
Range Sum Query - Immutable