最大和连续子数组的 Javascript 程序

编写一个高效的程序，在具有最大和的一维数字数组中找到连续子数组的和。

Kadane算法：

Initialize:
    max_so_far = INT_MIN
    max_ending_here = 0

Loop for each element of the array
  (a) max_ending_here = max_ending_here + a[i]
  (b) if(max_so_far < max_ending_here)
            max_so_far = max_ending_here
  (c) if(max_ending_here < 0)
            max_ending_here = 0
return max_so_far

解释：
Kadane 算法的简单思想是查找数组的所有正连续段（max_ending_here 用于此）。并跟踪所有正段之间的最大和连续段（max_so_far 用于此）。每次我们得到一个正和时，将它与 max_so_far 进行比较，如果它大于 max_so_far 则更新 max_so_far

Lets take the example:
    {-2, -3, 4, -1, -2, 1, 5, -3}

    max_so_far = max_ending_here = 0

    for i=0,  a[0] =  -2
    max_ending_here = max_ending_here + (-2)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=1,  a[1] =  -3
    max_ending_here = max_ending_here + (-3)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=2,  a[2] =  4
    max_ending_here = max_ending_here + (4)
    max_ending_here = 4
    max_so_far is updated to 4 because max_ending_here greater 
    than max_so_far which was 0 till now

    for i=3,  a[3] =  -1
    max_ending_here = max_ending_here + (-1)
    max_ending_here = 3

    for i=4,  a[4] =  -2
    max_ending_here = max_ending_here + (-2)
    max_ending_here = 1

    for i=5,  a[5] =  1
    max_ending_here = max_ending_here + (1)
    max_ending_here = 2

    for i=6,  a[6] =  5
    max_ending_here = max_ending_here + (5)
    max_ending_here = 7
    max_so_far is updated to 7 because max_ending_here is 
    greater than max_so_far

    for i=7,  a[7] =  -3
    max_ending_here = max_ending_here + (-3)
    max_ending_here = 4

程序：

Javascript

 
// JavaScript program to find maximum
// contiguous subarray
  
// Function to find the maximum
// contiguous subarray
function maxSubArraySum(a, size)
{
    var maxint = Math.pow(2, 53)
    var max_so_far = -maxint - 1
    var max_ending_here = 0
      
    for (var i = 0; i < size; i++)
    {
        max_ending_here = max_ending_here + a[i]
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here
 
        if (max_ending_here < 0)
            max_ending_here = 0
    }
    return max_so_far
}
  
// Driver code
var a = [ -2, -3, 4, -1, -2, 1, 5, -3 ]
document.write("Maximum contiguous sum is",
               maxSubArraySum(a, a.length))
  
// This code is contributed by AnkThon
 

输出：

Maximum contiguous sum is 7

另一种方法：

Javascript

        // JavaScript Program to implement
        // the above approach
 
        function maxSubarraySum(arr, size)
        {
            let max_ending_here = 0, max_so_far = Number.MIN_VALUE;
            for (let i = 0; i < size; i++) {
 
                // include current element to previous subarray only
                // when it can add to a bigger number than itself.
                if (arr[i] <= max_ending_here + arr[i]) {
                    max_ending_here += arr[i];
                }
 
                // Else start the max subarray from current element
                else {
                    max_ending_here = arr[i];
                }
                if (max_ending_here > max_so_far) {
                    max_so_far = max_ending_here;
                }
            }
            return max_so_far;
        }
         
// This code is contributed by Potta Lokesh
    

时间复杂度： O(n)
算法范式：动态规划
以下是Mohit Kumar建议的另一个简单实现。当数组中的所有数字都是负数时，该实现会处理这种情况。

Javascript

// C# program to print largest
// contiguous array sum
 
function maxSubArraySum(a,size)
{
  let max_so_far = a[0];
  let curr_max = a[0];
 
  for (let i = 1; i < size; i++)
  {
      curr_max = Math.max(a[i], curr_max+a[i]);
      max_so_far = Math.max(max_so_far, curr_max);
  }
 
return max_so_far;
}
 
// Driver code
 
let a = [-2, -3, 4, -1, -2, 1, 5, -3];
let n = a.length;
document.write("Maximum contiguous sum is ",maxSubArraySum(a, n));
     

输出：

Maximum contiguous sum is 7

为了打印具有最大和的子数组，我们在获得最大和时维护索引。

Javascript

// javascript program to print largest
// contiguous array sum   
function maxSubArraySum(a , size) {
        var max_so_far = Number.MIN_VALUE, max_ending_here = 0, start = 0, end = 0, s = 0;
 
        for (i = 0; i < size; i++) {
            max_ending_here += a[i];
 
            if (max_so_far < max_ending_here) {
                max_so_far = max_ending_here;
                start = s;
                end = i;
            }
 
            if (max_ending_here < 0) {
                max_ending_here = 0;
                s = i + 1;
            }
        }
        document.write("Maximum contiguous sum is " + max_so_far);
        document.write("<br/>Starting index " + start);
        document.write("<br/>Ending index " + end);
    }
 
    // Driver code
     
        var a = [ -2, -3, 4, -1, -2, 1, 5, -3 ];
        var n = a.length;
        maxSubArraySum(a, n);
 
// This code is contributed by Rajput-Ji

输出：

Maximum contiguous sum is 7
Starting index 2
Ending index 6

Kadane 的算法既可以被视为贪心算法，也可以被视为 DP。正如我们所看到的，我们正在保持整数的运行总和，当它小于 0 时，我们将其重置为 0（贪婪部分）。这是因为继续负总和比重新开始新范围更糟糕。现在它也可以被视为一个 DP，在每个阶段我们有 2 个选择：要么获取当前元素并继续之前的总和，要么重新开始一个新的范围。这两个选择都在实施过程中得到照顾。

时间复杂度： O(n)

辅助空间： O(1)

现在试试下面的问题
给定一个整数数组（可能某些元素为负数），编写一个 C 程序，通过将数组中 n ≤ ARRAY_SIZE 的“n”个连续整数相乘，找出*最大乘积*。另外，打印最大乘积子数组的起点。

有关更多详细信息，请参阅有关最大和连续子数组的完整文章！