在C ++中删除最多一个元素后，最大化最大子数组总和
问题陈述
给定一个N个整数的数组arr[]。任务是首先找到最大子数组总和，然后从子数组中移除最多一个元素。最多删除一个元素，以使删除后的最大和最大。
如果给定的输入数组为{1,2,3，-2,3}，则最大子数组总和为{2,3，-2,3}。然后我们可以删除-2。删除剩余数组后-
{1, 2, 3, 3} with sum 9 which is maximum.算法
1. Use Kadane’s algorithm to find the maximum subarray sum. 2. Once the sum has beens find, re-apply Kadane’s algorithm to find the maximum sum again with some minor changes)
示例
#include <bits/stdc++.h>
using namespace std;
int getMaxSubarraySum(int *arr, int n){
int max = INT_MIN;
int currentMax = 0;
for (int i = 0; i < n; ++i) {
currentMax = currentMax + arr[i];
if (max < currentMax) {
max = currentMax;
}
if (currentMax < 0) {
currentMax = 0;
}
}
return max;
}
int getMaxSum(int *arr, int n){
int cnt = 0;
int minVal = INT_MAX;
int minSubarr = INT_MAX;
int sum = getMaxSubarraySum(arr, n);
int max = INT_MIN;
int currentMax = 0;
for (int i = 0; i < n; ++i) {
currentMax = currentMax + arr[i];
++cnt;
minSubarr = min(arr[i], minSubarr);
if (sum == currentMax) {
if (cnt == 1) {
minVal = min(minVal, 0);
} else {
minVal = min(minVal, minSubarr);
}
}
if (currentMax < 0) {
currentMax = 0;
cnt = 0;
minSubarr = INT_MAX;
}
}
return sum - minVal;
}
int main(){
int arr[] = {1, 2, 3, -2, 3};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Maximum sum = " << getMaxSum(arr, n) << endl;
return 0;
}输出结果
当您编译并执行上述程序时。它生成以下输出-
Maximum sum = 9