My code:

public class Solution {
    public int maxSubArrayLen(int[] nums, int k) {
        if (nums == null || nums.length == 0)
            return 0;
        int max = 0;
        HashMap<Integer, Integer> tracker = new HashMap<Integer, Integer>();
        int sum = 0;
        tracker.put(0, -1);
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (!tracker.containsKey(sum)) {
                tracker.put(sum, i);
            }
            if (tracker.containsKey(sum - k)) {
                int pre = tracker.get(sum - k);
                max = Math.max(max, i - pre);
            }
        }
        return max;
    }
}

这道题目可以作为这类题目的通解。
找一段区域中数组的和，要达到某种要求，比如，等于k

参考网页：
http://www.cnblogs.com/EdwardLiu/p/5104280.html

Anyway, Good luck, Richardo!

My code:

import java.util.HashMap;

public class Solution {
    public int maxSubArrayLen(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        int n = nums.length;
        int[] sum = new int[n];
        int cum = 0;
        for (int i = 0; i < n; i++) {
            cum += nums[i];
            sum[i] = cum;
        }
        
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        map.put(k, -1);
        int max = 0;
        for (int i = 0; i < sum.length; i++) {
            if (map.containsKey(sum[i])) {
                int index = map.get(sum[i]);
                max = Math.max(max, i - index);
            }
            int target = sum[i] + k;
            if (!map.containsKey(target)) {
                map.put(sum[i] + k, i);
            }
        }
        
        return max;
    }
    
    public static void main(String[] args) {
        Solution test = new Solution();
        int[] nums = new int[]{1,-1,5,-2,3};
        int ret = test.maxSubArrayLen(nums, 3);
        System.out.println(ret);
    }
}

这道题目自己做了出来。
其实就是先构造一个累加数列。然后做一个 two sum

当我到达 sum[i]时，如果有 sum[j] + k = sum[i]
那么， [ j + 1, i ] 的和一定是 k
所以用一个 HashTable 用来存 target 的值。和 2Sum一个原理。

Anyway, Good luck, Richardo! — 09/21/2016