给定一个整数数组和一个整数key,如果子数组元素之和不小于key,返回该类子数组中长度最小子数组对应的
此题有两类解法:
1)暴力解法(使用两重for循环),代码如下所示
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int getMinLength(vector<int>& arr, int key) {
int minLength = arr.size();
for(int i = 0; i < arr.size(); i++) {
int sum = 0;
for(int j = i; j < arr.size(); j++) {
sum += arr[j];
if(sum >= key) {
int length = j - i + 1;
if(length == 1) {
return 1;
}
if(length < minLength) {
minLength = length;
}
}
}
}
return minLength;
}
int main() {
vector<int> arr;
string line;
string temp;
while (getline(cin, line)) {
int i;
for (i = 0; i < line.size(); i++) {
if(line[i] == '[') {
continue;
}
if(line[i] == ',') {
arr.emplace_back(stoi(temp));
temp.clear();
continue;
}
if(line[i] == ']') {
arr.emplace_back(stoi(temp));
temp.clear();
break;
}
temp += line[i];
}
int key = stoi(line.substr(i + 2));
int len = getMinLength(arr, key);
cout << len << endl;
arr.clear();
}
return 0;
}
算法的时间复杂度:O(n^2)
算法的空间复杂度:O(1)
2)滑动窗口法
类似于双指针法,慢指针指向子数组的头部,快指针指向子数组的尾部,当子数组之和大于key时,更新符合条件的子数组的最小长度,慢指针向后移动,当子数组之和小于key时,快指针向后移动,代码如下所示
int getMinLength(vector<int>& arr, int key) {
int minLength = arr.size();
int sum = 0;
int slowIndex = 0;
int fastIndex = 0;
sum += arr[fastIndex];
while(true) {
while(sum >= key) {
int length = fastIndex - slowIndex + 1;
if(length == 1) {
return 1;
}
if(length < minLength) {
minLength = length;
}
sum -= arr[slowIndex];
slowIndex++;
}
fastIndex++;
if(fastIndex < arr.size()) {
sum += arr[fastIndex];
}
else {
break;
}
}
return minLength;
}
算法的时间复杂度:O(n)
算法的空间复杂度:O(1)