325. Maximum Size Subarray Sum Eqauls K

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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?

解法1：

用prefix sum数组解决， 用一个map记录每一个accumulate sum对应的位置。然后线性扫描一遍加和数组，对于没有见过的数直接加入map，对于见过的就可以判断是否存在dp[i] - k, 注意。这题没有说是否有duplicate，实际是可能的，而我们push进map的都是第一次出现的位置，这样保证了如果找到一个解，一定是最长的一个subarray。
C++

1

Java

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public class Solution {
    public int maxSubArrayLen(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int n = nums.length;
        int[] dp = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            dp[i] = dp[i - 1] + nums[i - 1];
        }
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        map.put(0, 0);  // don't forget to put this into hashmap
        int max = 0;
        for (int i = 1; i <= n; i++) {
            int remain = dp[i] - k;
            if (map.containsKey(remain)) {
                max = Math.max(max, i - map.get(remain));
            }
            if (!map.containsKey(dp[i])) {
                map.put(dp[i], i);
            }
        }
        return max;
    }
}