[LeetCode] 325. Maximum Size Subarray Sum Equals k 和等于k的最长子数组

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;
    }
};

　　

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[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

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