问题陈述是:
Given an unsorted array of nonnegative integers, find a continous subarray which adds to a given number.
Examples:
Input: arr[] = {1, 4, 20, 3, 10, 5}, sum = 33
Ouptut: Sum found between indexes 2 and 4
Input: arr[] = {1, 4, 0, 0, 3, 10, 5}, sum = 7
Ouptut: Sum found between indexes 1 and 4
Input: arr[] = {1, 4}, sum = 0
Output: No subarray found
基于对来自 this post 的解决方案的解释以下解决方案不适用于负数:
/* An efficient program to print subarray with sum as given sum */
#include<stdio.h>
/* Returns true if the there is a subarray of arr[] with sum equal to 'sum'
otherwise returns false. Also, prints the result */
int subArraySum(int arr[], int n, int sum)
{
/* Initialize curr_sum as value of first element
and starting point as 0 */
int curr_sum = arr[0], start = 0, i;
/* Add elements one by one to curr_sum and if the curr_sum exceeds the
sum, then remove starting element */
for (i = 1; i <= n; i++)
{
// If curr_sum exceeds the sum, then remove the starting elements
while (curr_sum > sum && start < i-1)
{
curr_sum = curr_sum - arr[start];
start++;
}
// If curr_sum becomes equal to sum, then return true
if (curr_sum == sum)
{
printf ("Sum found between indexes %d and %d", start, i-1);
return 1;
}
// Add this element to curr_sum
if (i < n)
curr_sum = curr_sum + arr[i];
}
// If we reach here, then no subarray
printf("No subarray found");
return 0;
}
我尝试考虑了几种不同的输入场景,但我想不出输入数组包含负数而不会产生正确输出的情况。这是一个有效的负数输入数组:
int arr[] = { 15, 14, -2, 3, -5, 14};
当帖子说该解决方案不适用于负数时,指的是哪些输入案例?
最佳答案
这个解决方案依赖于这样一个事实,即当我们删除一个元素时,我们的总和会减少,但负数与这个假设相矛盾。
最短的例子是这样的情况
-1 5 2
当我们寻找和为5的子数组时,操作如下:
add -1, sum = -1
add 5, sum = 4
add 2, sum = 6
remove -1, sum = 7
remove 5, sum = 2
我们已经到了列表的末尾,但还没有找到所需的子数组。
关于c++ - 查找具有给定总和的子数组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39320735/