- GeeksforGeeks: Find subarray with given sum - GeeksforGeeks
Given an nonnegative integer array, find a subarray where the sum of numbers is k.
Your code should return the index of the first number and the index of the last number.
Example
Given [1, 4, 20, 3, 10, 5], sum k = 33, return [2, 4].
题 Zero Sum Subarray | Data Structure and Algorithm 的升级版，这道题求子串和为 K 的索引。首先我们可以考虑使用时间复杂度相对较低的哈希表解决。前一道题的核心约束条件为
#include <iostream>
#include <vector>
#include <map>
using namespace std;
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, int k){
vector<int> result;
// curr_sum for the first item, index for the second item
// unordered_map<int, int> hash;
map<int, int> hash;
hash[0] = 0;
int curr_sum = 0;
for (int i = 0; i != nums.size(); ++i) {
curr_sum += nums[i];
if (hash.find(curr_sum - k) != hash.end()) {
result.push_back(hash[curr_sum - k]);
result.push_back(i);
return result;
} else {
hash[curr_sum] = i + 1;
}
}
return result;
}
};
int main(int argc, char *argv[])
{
int int_array1[] = {1, 4, 20, 3, 10, 5};
int int_array2[] = {1, 4, 0, 0, 3, 10, 5};
vector<int> vec_array1;
vector<int> vec_array2;
for (int i = 0; i != sizeof(int_array1) / sizeof(int); ++i) {
vec_array1.push_back(int_array1[i]);
}
for (int i = 0; i != sizeof(int_array2) / sizeof(int); ++i) {
vec_array2.push_back(int_array2[i]);
}
Solution solution;
vector<int> result1 = solution.subarraySum(vec_array1, 33);
vector<int> result2 = solution.subarraySum(vec_array2, 7);
cout << "result1 = [" << result1[0] << " ," << result1[1] << "]" << endl;
cout << "result2 = [" << result2[0] << " ," << result2[1] << "]" << endl;
return 0;
}与 Zero Sum Subarray 题的变化之处有两个地方，第一个是判断是否存在哈希表中时需要使用hash.find(curr_sum - k), 最终返回结果使用result.push_back(hash[curr_sum - k]);而不是result.push_back(hash[curr_sum]);
略，见 Zero Sum Subarray | Data Structure and Algorithm
不知道细心的你是否发现这道题的隐含条件——nonnegative integer array, 这也就意味着子串和函数
我们首先来举个实际例子帮助分析，以整数数组 {1, 4, 20, 3, 10, 5} 为例，要求子串和为33的索引值。首先我们可以构建如下表所示的子串和
| 1 | 5 | 25 | 28 | 38 | |
|---|---|---|---|---|---|
| 0 | 1 | 2 | 3 | 4 |
要使部分子串和为33，则要求的第二个索引值必大于等于4，如果索引值再继续往后遍历，则所得的子串和必大于等于38，进而可以推断出索引0一定不是解。那现在怎么办咧？当然是把它扔掉啊！第一个索引值往后递推，直至小于33时又往后递推第二个索引值，于是乎这种技巧又可以认为是「两根指针」。
#include <iostream>
#include <vector>
#include <map>
using namespace std;
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> subarraySum2(vector<int> nums, int k){
vector<int> result;
int left_index = 0;
int curr_sum = 0;
for (int i = 0; i != nums.size(); ++i) {
curr_sum += nums[i];
if (curr_sum == k) {
result.push_back(left_index);
result.push_back(i);
return result;
}
while (curr_sum > k) {
curr_sum -= nums[left_index];
++left_index;
}
}
return result;
}
};
int main(int argc, char *argv[])
{
int int_array1[] = {1, 4, 20, 3, 10, 5};
int int_array2[] = {1, 4, 0, 0, 3, 10, 5};
vector<int> vec_array1;
vector<int> vec_array2;
for (int i = 0; i != sizeof(int_array1) / sizeof(int); ++i) {
vec_array1.push_back(int_array1[i]);
}
for (int i = 0; i != sizeof(int_array2) / sizeof(int); ++i) {
vec_array2.push_back(int_array2[i]);
}
Solution solution;
vector<int> result1 = solution.subarraySum2(vec_array1, 33);
vector<int> result2 = solution.subarraySum2(vec_array2, 7);
cout << "result1 = [" << result1[0] << " ," << result1[1] << "]" << endl;
cout << "result2 = [" << result2[0] << " ," << result2[1] << "]" << endl;
return 0;
}使用for循环累加curr_sum, 在curr_sum > k时再使用while递减curr_sum, 同时递增左边索引left_index.
看似有两重循环，由于仅遍历一次数组，且索引最多挪动和数组等长的次数。故最终时间复杂度近似为