算法题（如何计算两个数组中的最大求和路径（））

本文概述

C ++
Java
python
C#
的PHP

给定两个排序的数组, 这样该数组可能具有一些公共元素。查找从任何数组的开头到两个数组中的任意一个结尾的最大和路径的和。我们只能在公共元素处从一个阵列切换到另一个阵列。
注意：公用元素不必具有相同的索引。
预期时间复杂度：O(m + n), 其中m是ar1 []中的元素数, n是ar2 []中的元素数。
例子：

Input: ar1[] = {2, 3, 7, 10, 12} ar2[] = {1, 5, 7, 8} Output: 35Explanation: 35 is sum of 1 + 5 + 7 + 10 + 12. We start from the first element of arr2 which is 1, then we move to 5, then 7.From 7, we switch to ar1 (as 7 is common) and traverse 10 and 12.Input: ar1[] = {10, 12} ar2 = {5, 7, 9} Output: 22Explanation: 22 is the sum of 10 and 12. Since there is no common element, we need to take all elements from the array with more sum.Input: ar1[] = {2, 3, 7, 10, 12, 15, 30, 34} ar2[] = {1, 5, 7, 8, 10, 15, 16, 19} Output: 122Explanation: 122 is sum of 1, 5, 7, 8, 10, 12, 15, 30, 34

高效方法：这个想法是做一些类似的合并过程合并排序。这涉及计算两个数组的所有公共点之间的元素之和。只要有一个共同点, 就比较两个和并将两个的最大值相加。
算法:

创建一些变量, 结果, 总和1, 总和2。将结果初始化为0。还将两个变量sum1和sum2初始化为0。此处, sum1和sum2用于分别将元素的和存储在ar1 []和ar2 []中。这些总和在两个公共点之间。
现在运行一个循环以遍历两个数组的元素。遍历时, 按以下顺序比较数组1和数组2的当前元素。
1. 如果当前元素数组1小于的当前元素阵列2, 然后更新总和1, 否则, 如果当前元素为阵列2较小, 然后更新总和2.
2. 如果当前元素数组1和阵列2相同, 则取sum1和sum2的最大值, 并将其加到结果中。还将公共元素添加到结果中。
3. 这一步可以比作两个合并已排序数组, 如果要处理两个当前数组索引中最小的元素, 则可以保证如果有任何公共元素, 它将被一起处理, 因此可以处理两个公共元素之间的元素之和。

下面是上述代码的实现：
C ++

// C++ program to find maximum sum path #include < iostream> using namespace std; // Utility function to find maximum of two integers int max( int x, int y) { return (x > y) ? x : y; } // This function returns the sum of elements on maximum path // from beginning to end int maxPathSum( int ar1[], int ar2[], int m, int n) { // initialize indexes for ar1[] and ar2[] int i = 0, j = 0; // Initialize result and current sum through ar1[] and // ar2[]. int result = 0, sum1 = 0, sum2 = 0; // Below 3 loops are similar to merge in merge sort while (i < m & & j < n) { // Add elements of ar1[] to sum1 if (ar1[i] < ar2[j]) sum1 += ar1[i++]; // Add elements of ar2[] to sum2 else if (ar1[i] > ar2[j]) sum2 += ar2[j++]; else // we reached a common point { // Take the maximum of two sums and add to // result result += max(sum1, sum2); // Update sum1 and sum2 for elements after this // intersection point sum1 = 0, sum2 = 0; // Keep updating result while there are more // common elements int temp = i; while (i < m & & ar1[i] == ar2[j]) sum1 += ar1[i++]; while (j < n & & ar1[temp] == ar2[j]) sum2 += ar2[j++]; result += max(sum1, sum2); sum1 = 0, sum2 = 0; } } // Add remaining elements of ar1[] while (i < m) sum1 += ar1[i++]; // Add remaining elements of ar2[] while (j < n) sum2 += ar2[j++]; // Add maximum of two sums of remaining elements result += max(sum1, sum2); return result; } // Driver code int main() { int ar1[] = { 2, 3, 7, 10, 12, 15, 30, 34 }; int ar2[] = { 1, 5, 7, 8, 10, 15, 16, 19 }; int m = sizeof (ar1) / sizeof (ar1[0]); int n = sizeof (ar2) / sizeof (ar2[0]); // Function call cout < < "Maximum sum path is " < < maxPathSum(ar1, ar2, m, n); return 0; }

Java

// JAVA program to find maximum sum path class MaximumSumPath { // Utility function to find maximum of two integers int max( int x, int y) { return (x > y) ? x : y; } // This function returns the sum of elements on maximum // path from beginning to end int maxPathSum( int ar1[], int ar2[], int m, int n) { // initialize indexes for ar1[] and ar2[] int i = 0 , j = 0 ; // Initialize result and current sum through ar1[] // and ar2[]. int result = 0 , sum1 = 0 , sum2 = 0 ; // Below 3 loops are similar to merge in merge sort while (i < m & & j < n) { // Add elements of ar1[] to sum1 if (ar1[i] < ar2[j]) sum1 += ar1[i++]; // Add elements of ar2[] to sum2 else if (ar1[i] > ar2[j]) sum2 += ar2[j++]; // we reached a common point else { // Take the maximum of two sums and add to // result result += max(sum1, sum2); // Update sum1 and sum2 for elements after // this intersection point sum1 = 0 ; sum2 = 0 ; // Keep updating result while there are more // common elements int temp = i; while (i < m & & ar1[i] == ar2[j]) sum1 += ar1[i++]; while (j < n & & ar1[temp] == ar2[j]) sum2 += ar2[j++]; result += max(sum1, sum2); sum1 = 0 ; sum2 = 0 ; } } // Add remaining elements of ar1[] while (i < m) sum1 += ar1[i++]; // Add remaining elements of ar2[] while (j < n) sum2 += ar2[j++]; // Add maximum of two sums of remaining elements result += max(sum1, sum2); return result; } // Driver code public static void main(String[] args) { MaximumSumPath sumpath = new MaximumSumPath(); int ar1[] = { 2 , 3 , 7 , 10 , 12 , 15 , 30 , 34 }; int ar2[] = { 1 , 5 , 7 , 8 , 10 , 15 , 16 , 19 }; int m = ar1.length; int n = ar2.length; // Function call System.out.println( "Maximum sum path is :" + sumpath.maxPathSum(ar1, ar2, m, n)); } } // This code has been contributed by Mayank Jaiswal

python

# Python program to find maximum sum path # This function returns the sum of elements on maximum path from # beginning to end def maxPathSum(ar1, ar2, m, n): # initialize indexes for ar1[] and ar2[] i, j = 0 , 0 # Initialize result and current sum through ar1[] and ar2[] result, sum1, sum2 = 0 , 0 , 0 # Below 3 loops are similar to merge in merge sort while (i < m and j < n): # Add elements of ar1[] to sum1 if ar1[i] < ar2[j]: sum1 + = ar1[i] i + = 1 # Add elements of ar2[] to sum1 elif ar1[i] > ar2[j]: sum2 + = ar2[j] j + = 1 else :# we reached a common point # Take the maximum of two sums and add to result result + = max (sum1, sum2) # Update sum1 and sum2 for elements after this intersection point sum1, sum2 = 0 , 0 # Keep updating result while there are more common elements temp = i while i < m and ar1[i] = = ar2[j]: sum1 + = ar1[i]; i + = 1 while j< n and ar1[temp] = = ar2[j]: sum2 + = ar2[j] j + = 1 result + = max (sum1, sum2) sum1 = sum2 = 0 ; # Add remaining elements of ar1[] while i < m: sum1 + = ar1[i] i + = 1 # Add remaining elements of b[] while j < n: sum2 + = ar2[j] j + = 1 # Add maximum of two sums of remaining elements result + = max (sum1, sum2) return result # Driver code ar1 = [ 2 , 3 , 7 , 10 , 12 , 15 , 30 , 34 ] ar2 = [ 1 , 5 , 7 , 8 , 10 , 15 , 16 , 19 ] m = len (ar1) n = len (ar2) # Function call print "Maximum sum path is" , maxPathSum(ar1, ar2, m, n) # This code is contributed by __Devesh Agrawal__

// C# program for Maximum Sum Path in // Two Arrays using System; class GFG { // Utility function to find maximum // of two integers static int max( int x, int y) { return (x > y) ? x : y; } // This function returns the sum of // elements on maximum path from // beginning to end static int maxPathSum( int [] ar1, int [] ar2, int m, int n) { // initialize indexes for ar1[] // and ar2[] int i = 0, j = 0; // Initialize result and current // sum through ar1[] and ar2[]. int result = 0, sum1 = 0, sum2 = 0; // Below 3 loops are similar to // merge in merge sort while (i < m & & j < n) { // Add elements of ar1[] to sum1 if (ar1[i] < ar2[j]) sum1 += ar1[i++]; // Add elements of ar2[] to sum2 else if (ar1[i] > ar2[j]) sum2 += ar2[j++]; // we reached a common point else { // Take the maximum of two // sums and add to result result += max(sum1, sum2); // Update sum1 and sum2 for // elements after this // intersection point sum1 = 0; sum2 = 0; // Keep updating result while // there are more common // elements int temp = i; while (i < m & & ar1[i] == ar2[j]) sum1 += ar1[i++]; while (j < n & & ar1[temp] == ar2[j]) sum2 += ar2[j++]; result += max(sum1, sum2); sum1 = 0; sum2 = 0; } } // Add remaining elements of ar1[] while (i < m) sum1 += ar1[i++]; // Add remaining elements of ar2[] while (j < n) sum2 += ar2[j++]; // Add maximum of two sums of // remaining elements result += max(sum1, sum2); return result; } // Driver code public static void Main() { int [] ar1 = { 2, 3, 7, 10, 12, 15, 30, 34 }; int [] ar2 = { 1, 5, 7, 8, 10, 15, 16, 19 }; int m = ar1.Length; int n = ar2.Length; // Function call Console.Write( "Maximum sum path is :" + maxPathSum(ar1, ar2, m, n)); } } // This code is contributed by nitin mittal.

的PHP

< ?php // PHP Program to find Maximum Sum // Path in Two Arrays // This function returns the sum of // elements on maximum path // from beginning to end function maxPathSum( $ar1 , $ar2 , $m , $n ) {// initialize indexes for // ar1[] and ar2[] $i = 0; $j = 0; // Initialize result and // current sum through ar1[] // and ar2[]. $result = 0; $sum1 = 0; $sum2 = 0; // Below 3 loops are similar // to merge in merge sort while ( $i < $m and $j < $n ) {// Add elements of // ar1[] to sum1 if ( $ar1 [ $i ] < $ar2 [ $j ]) $sum1 += $ar1 [ $i ++]; // Add elements of // ar2[] to sum2 else if ( $ar1 [ $i ] > $ar2 [ $j ]) $sum2 += $ar2 [ $j ++]; // we reached a // common point else {// Take the maximum of two // sums and add to result $result += max( $sum1 , $sum2 ); // Update sum1 and sum2 for // elements after this // intersection point $sum1 = 0; $sum2 = 0; // Keep updating result while // there are more common // elements $temp = $i ; while ( $i < $m & & $ar1 [ $i ] == $ar2 [ $j ]) $sum1 += $ar1 [ $i ++]; while ( $j < $n & & $ar1 [ $temp ] == $ar2 [ $j ]) $sum2 += $ar2 [ $j ++]; $result += max( $sum1 , $sum2 ); $sum1 = 0; $sum2 = 0; } } // Add remaining elements of ar1[] while ( $i < $m ) $sum1 += $ar1 [ $i ++]; // Add remaining elements of ar2[] while ( $j < $n ) $sum2 += $ar2 [ $j ++]; // Add maximum of two sums // of remaining elements $result += max( $sum1 , $sum2 ); return $result ; } // Driver Code $ar1 = array (2, 3, 7, 10, 12, 15, 30, 34); $ar2 = array (1, 5, 7, 8, 10, 15, 16, 19); $m = count ( $ar1 ); $n = count ( $ar2 ); // Function call echo "Maximum sum path is " , maxPathSum( $ar1 , $ar2 , $m , $n ); // This code is contributed by anuj_67. ?>

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