The maximum, minimum, and absolute value of the sum of successive sub-arrays, the minimum value

#include <iostream> #include <vector> #include <algorithm>using namespace std;//the smallest and/or use of the DP idea , loop through each element in the array, add them together, if it is greater than 0, then//the sum of the current element is cleared to 0, otherwise, and the minimum and comparison, update the minimum and, finally get must be sub-array of minimum and/or time complexity: O (n) space complexity: O (1) int minsum (vector <int> &num) {int min_sum = 0, sum = 0;for (int i=0; i<num.size (); i++) {sum + = num[i];if (sum > 0) sum = 0;if ( Sum < min_sum) min_sum = sum;} if (min_sum = = 0)//array only positive number {min_sum = num[0];for (int i=1; i<num.size (); i++) {if (num[i]<min_sum) min_sum = Num[i];}} return min_sum;} To find the maximum and/or use of the sub-array is the idea of DP, sequentially iterate through each element in the array, add them together, if the sum of less than 0, then//the current element is cleared to 0, otherwise, and the maximum and comparison, update the maximum and, finally get must be the maximum and/or time complexity of the subarray: Maxsum (vector<int> &num) {int max_sum = 0, sum = 0;for (int i=0; i<num.size (); i++) {sum + = num[i];if (Sum < 0 sum = 0;if (Sum > max_sum) max_sum = sum;} if (max_sum = = 0)//array only negative numbers {max_sum = num[0];for (int i=1; i<num.size (); i++) {if (num[i]>max_sum) max_sum = Num[i];}} return max_sum;} Find the minimum value of the sum of the subarray//violence-poor lift//time complexity: O (n^2) spatial complexity: O (1) int minAbsSum1 (vector<Int> &num) {int min_abs_sum = int_max;for (int i=0; i<num.size (); i++) {int cur_sum = 0;for (int j=i; J<num.size ( ); J + +) {cur_sum + = Num[j];if (ABS (Cur_sum) < min_abs_sum) Min_abs_sum = ABS (Cur_sum);}} return min_abs_sum;} Find the minimum value of the sum of the subarray//Calculate all Sum[0-j] 0<= J <n, then sort the array of sum[0-j], then for any i,j segment and equal to: sum[i-j]= Sum[0-j]-sum[0-i];// Set the array sum to store the 0-j of the sub-array and//Because the sum has been sorted, only the minimum value of Sum[z]-sum[z-1],sum[z] (0<=z<sum.size ()) must be found after sorting. Z is the sorted index//If the sum[z] case, Z is the sorted index, then Minabs = ABS (Sum[0-i])//If it is sum[z]-sum[z-1] case, then Minabs = ABS (Sum[i]-sum[j])//time complexity: O (NLOGN) Spatial complexity: O (n) int minAbsSum2 (vector<int> &num) {if (Num.size () ==0) return 0;if (Num.size () ==1) return ABS ( NUM[0]); int min_abs_sum;vector<int> Sum;int cur_sum = 0;for (int i=0; i<num.size (); i++) {cur_sum + = Num[i];if ( Cur_sum = = 0) return 0;sum.push_back (cur_sum);} Sort (Sum.begin (), Sum.end ()), Min_abs_sum = ABS (Sum[0]), for (int i=0; i<sum.size ()-1; i++) {int TEMP1 = ABS (sum[i+1]); int TEMP2 = ABS (Sum[i+1]-sum[i]); cur_sum =(TEMP1&LT;TEMP2) temp1:temp2;if (cur_sum = = 0) return 0;if (cur_sum<min_abs_sum) min_abs_sum = cur_sum;} return min_abs_sum;} Find the maximum da value of the sum of the subarray//violent poor lift//time complexity: O (n^2) spatial complexity: O (1) int maxAbsSum1 (vector<int> &num) {int max_abs_sum = 0;for ( int i=0; I<num.size (); i++) {int cur_sum = 0;for (int j=i; j<num.size (); j + +) {cur_sum + = Num[j];if (ABS (cur_sum) > max_abs_sum) max_abs_sum = a BS (cur_sum);}} return max_abs_sum;} Find the maximum value of the sum of the subarray//Calculate all Sum[0-j] 0<= J <n, then sort the array of sum[0-j], then for any i,j segment and equal to: sum[i-j]= Sum[0-j]-sum[0-i];// Set the array sum to store the 0-j of the sub-array and//Because the sum has been sorted, only the maximum value of sum[sum.size () -1]-sum[0],sum[z] (0<=z<sum.size ()) must be found after sorting. Z is the sorted index//if SUM[Z], Z is the sorted index, then Maxabs = ABS (Sum[0-i])//If it is sum[sum.size () -1]-sum[0] case, then Maxabs = ABS (Sum[i]-sum[j]) Complexity of Time: O (nlogn) spatial complexity: O (n) int maxAbsSum2 (vector<int> &num) {if (Num.size () ==0) return 0;if (Num.size () ==1) Return ABS (Num[0]), int max_abs_sum;vector<int> sum;int cur_sum = 0;for (int i=0; i<num.size (); i++) {cur_sum + = Nu M[i];iF (cur_sum = = 0) return 0;sum.push_back (cur_sum);} Sort (Sum.begin (), Sum.end ()), Max_abs_sum = ABS (Sum[sum.size () -1]-sum[0]); for (int i=0; i<sum.size (); i++) {cur_sum = ABS (Sum[i]); if (cur_sum>max_abs_sum) max_abs_sum = cur_sum;} return max_abs_sum;} int main () {int n;while (cin>>n) {vector<int> num;int number;for (int i=0; i<n; i++) {cin>>number; Num.push_back (number);} cout<< "minsum =" <<minsum (num) <<endl;cout<< "maxsum =" <<maxsum (num) <<endl;cout << "MINABSSUM1 =" <<minabssum1 (num) <<endl;cout<< "minAbsSum2 =" <<minabssum2 (num) < <endl;cout<< "MAXABSSUM1 =" <<maxabssum1 (num) <<endl;cout<< "maxAbsSum2 =" << MAXABSSUM2 (num) <<endl;cout<< "------------------------------------" &LT;&LT;ENDL;} return 0;}

The maximum, minimum, and absolute value of the sum of successive sub-arrays, the minimum value