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].
题解1 – 哈希表
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]
源码分析
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]);
复杂度分析
略，见 Zero Sum Subarray | Data Structure and Algorithm
题解2 – 利用单调函数特性
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).
| f(i) | 1 | 5 | 25 | 28 | 38 |
|---|---|---|---|---|---|
| i | 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」。
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]
源码分析
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)。
复杂度分析
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