数据结构(8) -- 算法应用实例

文章目录

1.最大子列和问题

算法1:

public class Demo5 {
    static int[] list = {-2, 11, -4, 13, -5, -2};
    //算法1：
    public static int maxSubseqSum1(int[] list) {
        int length = list.length;
        int i, j, k;
        int thisSum, maxSum = 0;
        for (i = 0; i < length; i++) {  //i是子列子列左端位置
            for (j = i; j < length; j++) { //j是子列右端位置
                thisSum = 0; //thisSum是list[i]到list[j]的子列和
                for (k = i; k <= j; k++) {
                    thisSum += list[k];
                    if (thisSum > maxSum) {
                        maxSum = thisSum;
                    }
                }
            }
        }
        return maxSum;
    }

    public static void main(String[] args) {
        int i = maxSubseqSum1(list);
        System.out.println(i);
    }
}

运行结果:

算法2：

//算法2：
    public static int maxSubseqSum2(int[] list) {
        int length = list.length;
        int i, j, k;
        int thisSum, maxSum = 0;
        for (i = 0; i < length; i++) {  //i是子列子列左端位置
            thisSum = 0; //thisSum是list[i]到list[j]的子列和
            for (j = i; j < length; j++) { //j是子列右端位置
                thisSum += list[j ];
                if (thisSum > maxSum) { //如果刚得到的这个子列和更大
                    maxSum = thisSum;  //则更新结果
                }
            }
        }
        return maxSum;
    }
      public static void main(String[] args) {
        int sum1 = maxSubseqSum1(list);
        int sum2 = maxSubseqSum1(list);
        System.out.println("算法1;" + sum1);
        System.out.println("算法2;" + sum2);
    }

算法3：

但是这个仍不是最快的算法.

算法4,在线处理:

//算法4：在线处理
/算法4：在线处理
    public static int maxSubseqSum4(int[] list) {
        int thisSum, maxSum;
        thisSum = maxSum = 0;
        for (int i = 0; i < list.length; i++) {
            thisSum += list[i]; //向右累加
            if (thisSum > maxSum) {
                maxSum = thisSum; //发现更大和则更新当前结果
            }else if (thisSum < 0) {//如果当前子列和为负
                thisSum = 0; //则不可能使后面的部门和增大，抛弃之。
            }
        }
        return maxSum;
    }

总结:

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完