连续子数组的最大和

来源：互联网发布：2014网络歌曲大全编辑：程序博客网时间：2024/06/05 05:30

题目描述

HZ偶尔会拿些专业问题来忽悠那些非计算机专业的同学。今天测试组开完会后,他又发话了:在古老的一维模式识别中,常常需要计算连续子向量的最大和,当向量全为正数的时候,问题很好解决。但是,如果向量中包含负数,是否应该包含某个负数,并期望旁边的正数会弥补它呢？例如:{6,-3,-2,7,-15,1,2,2},连续子向量的最大和为8(从第0个开始,到第3个为止)。你会不会被他忽悠住？

思路大概有3种。

暴力穷举法
分治递归
动态规划

import java.util.Scanner;public class FindGreatestSumOfSubArray {    public static void main(String[] args) {        Scanner scanner = new Scanner(System.in);        System.out.println("the size of array:");        int n = scanner.nextInt();        int[] array = new int[n];        System.out.println("input the array:");        for (int i = 0; i < n; i++) {            array[i] = scanner.nextInt();        }        System.out.println("max sumOfSubArray is:" + FindGreatestSumOfSubArray(array));    }    // 方法1：暴力穷举法    public static int FindGreatestSumOfSubArray(int[] array) {        int len = array.length;        if (len == 0) {            return 0;        }        int maxSum = array[0];        for (int i = 0; i < len - 1; i++) {            int partSum = array[i];            if (partSum > maxSum) {                maxSum = partSum;            }            for (int j = i + 1; j < len; j++) {                partSum += array[j];                if (partSum > maxSum) {                    maxSum = partSum;                }            }        }        return maxSum;    }    // 方法2：分治递归方法 o(N*log2N)    public static int FindGreatestSumOfSubArray1(int[] array) {        int len = array.length;        if (len == 0) {            return 0;        }        return maxSumOfSubArray(array, 0, len - 1);    }    // 递归求解    public static int maxSumOfSubArray(int[] array, int left, int right) {        if (left == right) {            return array[left];        }        int mid = (left + right) / 2;        // 前半部分最大子数组        int leftMaxSub = maxSumOfSubArray(array, left, mid);        // 后半部分最大子数组        int rightMaxSub = maxSumOfSubArray(array, mid + 1, right);        // 中点向左最大子数组        int maxOfMidLeft = maxSubArray(array, mid, left);        // 中点向右最大子数组        int maxOfMidRight = maxSubArray(array, mid + 1, right);        int midMaxSub = maxOfMidLeft + maxOfMidRight;        // 返回3种情况的最大值        int maxSub = leftMaxSub > rightMaxSub ? leftMaxSub : rightMaxSub;        return maxSub > midMaxSub ? maxSub : midMaxSub;    }    // 求left到right的最大子数组的最大值    public static int maxSubArray(int[] array, int left, int right) {        int maxSub = array[left];        if (left == right) {            return maxSub;        }        int tempSub = maxSub;        int flag = left < right ? 1 : -1;        if (flag == 1) {            for (int i = left + 1; i <= right; i++) {                tempSub += array[i];                if (tempSub > maxSub) {                    maxSub = tempSub;                }            }        } else {            for (int i = left - 1; i >= right; i--) {                tempSub += array[i];                if (tempSub > maxSub) {                    maxSub = tempSub;                }            }        }        return maxSub;    }    // 动态规划 O(N)复杂度    // all[i-1] =    // max{array[i-1],array[i-1]+Start[i],all[i]},其中Start[i]是指array[i],...,array[len-1]中从array[i]开始的最大子数组    // all[i]指array[i],...,array[len-1]中的最大子数组之和    public static int FindGreatestSumOfSubArray2(int[] array) {        int len = array.length;        if (len == 0) {            return 0;        }        int[] all = new int[len];        int[] start = new int[len];        all[len - 1] = array[len - 1];        start[len - 1] = array[len - 1];        for (int i = len - 2; i >= 0; i--) {            start[i] = Math.max(start[i + 1] + array[i], array[i]);            all[i] = Math.max(start[i], all[i + 1]);        }        return all[0];    }    // 动态规划的改进 o(1)的空间    public static int FindGreatestSumOfSubArray3(int[] array) {        int len = array.length;        if (len == 0) {            return 0;        }        int all = array[len - 1];        int start = array[len - 1];        for (int i = len - 2; i >= 0; i--) {            start = Math.max(start + array[i], array[i]);            all = Math.max(start, all);        }        return all;    }}

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