[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 = ;
         has[] = -;
         vector<int> res;
         for (int i = ; i < nums.size(); ++i) {
             sum += nums[i];
             if (has.find(sum) != has.end()) {
                 res.push_back(has[sum] + );
                 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(), );
         for (int i = ; i < acc.size(); ++i) {
             acc[i] += acc[i-];
         }
         int tmp, res = ;
         for (int i = ; i < acc.size() - ; ++i) {
             for (int j = i + 1; j < acc.size(); ++j) {
                 tmp = acc[j] - acc[i];
                 if (tmp>= start && tmp <= end) ++res;
             }
         }
         return res;
     }
 };