求最大字段和

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问题描述:

给定任意n个整数(可能为负数),组成序列a1,a2,a3,a4,a5,...,an, 求该序列的的子段和的最大值.

输入:

-2,11,-4,13,-5,-2

输出:

最大子段为: 11,-4,13 结果为:11-4+13=20

第一种方法,暴力求解:

package spam;/** * Created by panther on 17-3-22. */public class MaxSubSegment {    public static void main(String[] args) {        int arr[] = new int[]{-2, 11, -4, 13, -5, -2};        System.out.println(maxSum(arr.length, arr));    }    private static int maxSum(int n, int[] arr) {        int sum = 0;        for (int i = 0; i < n; ++i) {            for (int j = i; j < n; ++j) {                int thissum = 0;                for (int k = i; k < j; ++k) {                    thissum += arr[k];                }                if (thissum > sum) {                    sum = thissum;                }            }        }        return sum;    }}

时间复杂度为: O(n^3)

算法优化:

package spam;/** * Created by panther on 17-3-22. */public class MaxSubSegment {    public static void main(String[] args) {        int arr[] = new int[]{-2, 11, -4, 13, -5, -2};        System.out.println(maxSum(arr.length, arr));    }    private static int maxSum(int n, int[] arr) {        int sum = 0;        for (int i = 0; i < n; ++i) {            int thissum = 0;            for (int j = i; j < n; ++j) {//                for (int k = i; k < j; ++k) {                thissum += arr[j];//                }                if (thissum > sum) {                    sum = thissum;                }            }        }        return sum;    }}

改进后的算法少了一层循环,避免了求和的重复计算,时间复杂度为O(n^2)

动态规划算法实现:

记b[j]为从1到j的最大字段和,由定义可知道,当b[j-1]>0时, b[j]=b[j-1]+a[j],否则b[j]=a[j].

由此可以知道b[j]的动态规划递归公式:

b[j]=max{b[j-1]+a[j],a[j]},1<=j<=n

算法实现

private static int dynamicMaxSum(int n, int[] arr) {        int sum = 0, b = 0;        for (int i = 0; i < n; ++i) {            if (b > 0) {                b += arr[i];            } else {                b = arr[i];            }            if (b > sum) sum = b;        }        return sum;    }

时间复杂度是O(n)

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