【算法导论】最大子数组——暴力求解

最新推荐文章于 2021-11-27 13:14:25 发布

转载最新推荐文章于 2021-11-27 13:14:25 发布 · 524 阅读

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原文链接：https://yq.aliyun.com/articles/653968

文章标签：

#python #java #c/c++

本文介绍了使用暴力方法解决最大子数组和问题的具体实现过程。分别提供了 Java、Python 和 C 语言的实现代码，并详细展示了如何通过双重循环遍历数组的所有子数组来找到具有最大和的子数组。

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1. 暴力方法求解最大子数组问题：求出所有子数组的和并比较；

2. 伪代码

FIND-MAXIMUM-SUBARRAY(A)
    max = -∞
    for i = 1 to A.length
        sum = 0
        for j = i to A.length
            sum += A[j]
            if sum > max
                max = sum
                low = i
                high = j
    return (low, high, max)

3. 代码实现

java

public class MaxArray {

    private static class Result {
        int low;
        int high;
        int sum;

        public Result(int low, int high, int sum) {
            this.low = low;
            this.high = high;
            this.sum = sum;
        }
    }

    static Result findMaximumSubarray(int[] A){
        int max = Integer.MIN_VALUE;
        int low = 0;
        int high = 0;
        for (int i = 0; i < A.length; i++) {
            int sum = 0;
            for (int j = i; j < A.length; j++) {
                sum += A[j];
                if (sum > max) {
                    max = sum;
                    low = i;
                    high = j;
                }
            }
        }
        return new Result(low, high, max);
    }
}

python

之前用切片其实也是暴力求解

def find_maximum_subarray1(nums):
    max_sum = -float('inf')
    result = {}
    for i in range(len(nums)):
        total = 0
        for j in range(i, len(nums)):
            print(j)
            total += nums[j]
            if total > max_sum:
                max_sum = total
                result["low"] = i
                result["high"] = j
    result["sum"] = max_sum
    return result

C语言

typedef struct {
    int low;
    int high;
    int sum;
} result;

result find_maximum_subarray(int arr[], int len)
{
    int max = -((unsigned)(~0) >> 1);
    int low, high, i, j;
    for (i = 0; i < len; i++)
    {
        int sum = 0;
        for(j = i; j < len; j++)
        {
            sum += arr[j];
            if (sum > max)
            {
                max = sum;
                low = i;
                high = j;
            }
        }
    }
    result re;
    re.low = low;
    re.high = high;
    re.sum = max;
    return re;
}