LintCode 911: Maximum Size Subarray Sum Equals k_911. maximum size subarray sum equals k-优快云博客

本文介绍了解决寻找数组中最大长度的子数组，其元素和等于给定值k的问题。通过使用前缀和与哈希表，或者前缀和与双指针的方法，实现了O(n)时间复杂度的高效解决方案。

911. Maximum Size Subarray Sum Equals k

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

Example1

Input: nums = [1, -1, 5, -2, 3], k = 3

Output: 4

Explanation:

because the subarray [1, -1, 5, -2] sums to 3 and is the longest.

Example2

Input: nums = [-2, -1, 2, 1], k = 1

Output: 2

Explanation:

because the subarray [-1, 2] sums to 1 and is the longest.

Challenge

Can you do it in O(n) time?

Notice

The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.

解法1：presums + hashmap

先构建presums数组，然后就可以转换成一个数组里面找两个数之差等于k的问题了。

这个问题可以用同向双指针，也可以用hashmap。

注意，我们要考虑某个数本身就是k和某个presum本身就是k这两种特殊情况，因为presums里面两个元素相减的差并不会考虑头一个元素，所以会漏掉presum的情况。

class Solution {
public:
    /**
     * @param nums: an array
     * @param k: a target value
     * @return: the maximum length of a subarray that sums to k
     */
    int maxSubArrayLen(vector<int> &nums, int k) {
        int n = nums.size();
        if (n == 0) return 0;
        vector<int> presums(n);
        int maxLen = 0;
        
        presums[0] = nums[0];
        if (nums[0] == k) maxLen = 1;
        for (int i = 1; i < n; ++i) {
            presums[i] = presums[i - 1] + nums[i];
            if (nums[i] == k) maxLen = max(maxLen, 1);
            if (presums[i] == k) maxLen = max(maxLen, i + 1);
        }
        unordered_map<int, int> hashMap; //value, index
        for (int i = 0; i < n; ++i) {
            int temp = presums[i] - k;
            if (hashMap.find(temp) != hashMap.end()) {
                maxLen = max(maxLen, i - hashMap[temp]);
            }
            if (hashMap.find(presums[i]) == hashMap.end()) hashMap[presums[i]] = i;
        }

        return maxLen;
    }
};

解法2：presumes + two pointer (同向)

注意！这题不能用同向双指针(滑动窗口)！因为测试用例里面有负数！下面这个链接讲的很好。
算法学习-滑动窗口_滑动窗口处理-优快云博客