Subarray Sum K详解编程语言

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 f(i1)−f(i2)=0，这道题则变为 f(i1)−f(i2)=k

C++:

#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; 

}

输出：

result1 = [2 ,4] 

result2 = [1 ,4]

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 f(i) 为「单调不减」函数。单调函数在数学中可是重点研究的对象，那么如何将这种单调性引入本题中呢？不妨设 i2 i1, 题中的解等价于寻找 f(i2)−f(i1)=k, 则必有 f(i2)≥k.

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 f(i).

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」。

C++：

#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, curr_sum = 0; 

 for (int i = 0; i != nums.size(); ++i) { 

 while (curr_sum k) { 

 curr_sum -= nums[left_index]; 

 ++left_index; 

 if (curr_sum == k) { 

 result.push_back(left_index); 

 result.push_back(i - 1); 

 return result; 

 curr_sum += nums[i]; 

 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; 

}

输出：

result1 = [2 ,4] 

result2 = [1 ,4]

使用for循环, 在curr_sum k时使用while递减curr_sum, 同时递增左边索引left_index, 最后累加curr_sum。如果顺序不对就会出现 bug, 原因在于判断子串和是否满足条件时在递增之后(谢谢 @glbrtchen 汇报 bug)。

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 O(2n), 空间复杂度为 O(1).

原创文章，作者：ItWorker，如若转载，请注明出处：https://blog.ytso.com/20674.html