JAVA算法：求连续子数组的最大和（详细讲解）
JAVA算法：求连续子数组的最大和（详细讲解）
给定一个一维整型数组，请这个数组中连续子数组的最大和。
这个题目很常见，可以参考我的另外两篇博文
https://blog.csdn.net/seagal890/article/details/88616329
https://blog.csdn.net/seagal890/article/details/89062176
这里说明一下 Kadane 算法（Kadane’s Algorithm）
伪代码如下：
Initialize:
max_so_far = 0
max_ending_here = 0
Loop for each element of the array
(a) max_ending_here = max_ending_here + a[i]
(b) if(max_ending_here < 0)
max_ending_here = 0
(c) if(max_so_far < max_ending_here)
max_so_far = max_ending_here
return max_so_far算法解释：
Kadane算法的简单思想是查找数组的所有正的连续段（这里使用max_ending_here）。并跟踪所有正段之间的最大和连续段（使用max_so_far）。每次我们得到一个正和，将它与max_so_far比较，如果它大于max_so_far，则更新max_so_far。
例如：给定数组 {-2, -3, 4, -1, -2, 1, 5, -3}
初始时，令：
max_so_far = max_ending_here = 0
当 i=0时，a[0] = -2
max_ending_here = max_ending_here + (-2)
则：max_ending_here = 0 ，因为 max_ending_here < 0
当 i=1时，a[1] = -3
max_ending_here = max_ending_here + (-3)
则：max_ending_here = 0，因为 max_ending_here < 0
当 i=2时，a[2] = 4
max_ending_here = max_ending_here + (4)
max_ending_here = 4
则：max_so_far 更新为4，因为 max_ending_here =4 大于 max_so_far = 0
当 i=3时，a[3] = -1
max_ending_here = max_ending_here + (-1)
max_ending_here = 3
当 i=4时，a[4] = -2
max_ending_here = max_ending_here + (-2)
max_ending_here = 1
当 i=5时，a[5] = 1
max_ending_here = max_ending_here + (1)
max_ending_here = 2
当 i=6时，a[6] = 5
max_ending_here = max_ending_here + (5)
max_ending_here = 7
则：max_so_far 更新为 7，因为此时 max_ending_here 比 max_so_far 大
当 i=7时，a[7] = -3
max_ending_here = max_ending_here + (-3)
max_ending_here = 4
则最终连续子数组的最大和为：7，这就是答案。
package com.bean.algorithm.basic;
public class MaxSumofCountinousSubarray {
public static void main(String[] args) {
int[] a = { -2, -3, 4, -1, -2, 1, 5, -3 };
System.out.println("Maximum contiguous sum is " + maxSubArraySum(a));
}
static int maxSubArraySum(int a[]) {
int size = a.length;
int max_so_far = Integer.MIN_VALUE;
int max_ending_here = 0;
for (int 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;
}
}
程序运行结果：
Maximum contiguous sum is 7
优化后的算法设计：
static int maxSubArraySum(int a[]) {
int max_so_far = 0;
int max_ending_here = 0;
int size = a.length;
for (int i = 0; i < size; i++) {
max_ending_here = max_ending_here + a[i];
if (max_ending_here < 0)
max_ending_here = 0;
/*
* Compare only when max_ending_here > 0
*/
else if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}程序运行结果：
Maximum contiguous sum is 7
下面的算法增加了列出最大连续子数组和的起始下标和结束下标：
package com.bean.algorithm.basic;
public class MaxSumofCountinousSubarray2 {
static void maxSubArraySum(int a[], int size) {
int max_so_far = Integer.MIN_VALUE, max_ending_here = 0, start = 0, end = 0, s = 0;
for (int 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;
}
}
System.out.println("Maximum contiguous sum is " + max_so_far);
System.out.println("Starting index " + start);
System.out.println("Ending index " + end);
}
public static void main(String[] args) {
int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };
int n = a.length;
maxSubArraySum(a, n);
}
}
程序运行结果：
Maximum contiguous sum is 7
Starting index 2
Ending index 6