[LintCode] Subarray Sum & Subarray Sum II

Subarray Sum

Given an integer array, find a subarray where the sum of numbers is zero. Your code should return the index of the first number and the index of the last number.

Example

Given [-3, 1, 2, -3, 4], return [0, 2] or [1, 3].

用hashmap来存从nums[0]到nums[i]的和以及当前下标i，遍历数组求前缀和acc[i]，如果发现acc[i]的值已经在hashmap中的，那么说明已经找到一段子数组和为0了。

class Solution {
public:
    /**
     * @param nums: A list of integers
     * @return: A list of integers includes the index of the first number 
     *          and the index of the last number
     */
    vector<int> subarraySum(vector<int> nums){
        // write your code here
        map<int, int> has;
        int sum = 0;
        has[0] = -1;
        vector<int> res;
        for (int i = 0; i < nums.size(); ++i) {
            sum += nums[i];
            if (has.find(sum) != has.end()) {
                res.push_back(has[sum] + 1);
                res.push_back(i);
                return res;
            } else {
                has[sum] = i;
            }
        }
        return res;
    }
};

Subarray Sum II 

Given an integer array, find a subarray where the sum of numbers is between two given interval. Your code should return the number of possible answer.

Example

Given [1,2,3,4] and interval = [1,3], return 4. The possible answers are:

[0, 0]
[0, 1]
[1, 1]
[3, 3]

先求出前缀和，然后枚举求两个前缀和的差。如果差在start与end之间，就给res+1。注意前缀和数组前要插入一个0。

class Solution {
public:
    /**
     * @param A an integer array
     * @param start an integer
     * @param end an integer
     * @return the number of possible answer
     */
    int subarraySumII(vector<int>& A, int start, int end) {
        // Write your code here
        vector<int> acc(A);
        acc.insert(acc.begin(), 0);
        for (int i = 1; i < acc.size(); ++i) {
            acc[i] += acc[i-1];
        }
        int tmp, res = 0;
        for (int i = 0; i < acc.size() - 1; ++i) {
            for (int j = i + 1; j < acc.size(); ++j) {
                tmp = acc[j] - acc[i];
                if (tmp>= start && tmp <= end) ++res;
            }
        }
        return res;
    }
};