分治算法(divide and conquer)
分治算法由两部分组成:
1.分(divide):递归将一个大的问题分解为2个或2个以上的相同小问题并对子问题求解。
2.治(conquer): 从若干个子问题的解构建原问题的解。
使用分治法时应注意以下两点:
1.将一个大的问题分解成为若干小问题时应该是相同的若干小问题。
2.分解的若干子问题中,不应该包含其它子问题。也就是不要在若干个子问题中去求一个或多个想同的子问题的解。若算法中存在这样的问题时间效率将大幅度下降。
最大子序列求和例子:
import java.util.Random;
import java.util.Scanner;
/**
* 分治算法--最大子序列求和问题
*
* @分(divide):递归解决较小的问题
* @治(conquer):然后从子问题的解构建原问题的解
* @说明:一般坚持子问题是不相交的(即基本上不重叠)
* @author lls
*
*/
public class DivideAndConquerAlgorithm {
public static void main(String[] args) {
init();// 数据初始化
}
private static void init() {
Scanner sc = new Scanner(System.in);
Random rand = new Random();
int n = sc.nextInt();
int[] array = new int[n];
for (int i = 0; i < n; i++) {
array[i] = rand.nextInt(10) - rand.nextInt(10);
System.out.print(array[i] + " ");
}
System.out.println();
System.out.println("maxSubSum3=" + maxSubSum3(array));
System.out.println("maxSubSum4=" + maxSubSum4(array));
sc.close();
}
/**
* 本算法是为一个更为简单有效的求解最大子序列和的方法
* 时间效率为0(n)
*/
public static int maxSubSum4(int[] a) {
int maxSum = 0, thisSum = 0;
for (int j = 0; j < a.length; j++) {
thisSum += a[j];
if (thisSum > maxSum)
maxSum = thisSum;
else if (thisSum < 0)
thisSum = 0;
}
return maxSum;
}
/**
* Driver for divide-and-conquer maximum contiguous subsequence sum algorithm.
*/
public static int maxSubSum3(int[] a) {
return maxSumRec(a, 0, a.length - 1);
}
/**
* Recursive maximum contiguous subsequence sum algorithm.
* Finds maximum sum in subarray spanning a[left....right].
* Does not attempt to maintain actual best sequence.
*/
private static int maxSumRec(int[] a, int left, int right) {
if (left == right) // Base case
if (a[left] > 0)
return a[left];
else
return 0;
int center = (left + right) / 2;
int maxLeftSum = maxSumRec(a, left, center);
int maxRightSum = maxSumRec(a, center + 1, right);
int maxLeftBorderSum = 0, leftBorderSum = 0;
for (int i = center; i >= left; i--) {
leftBorderSum += a[i];
if (leftBorderSum > maxLeftBorderSum)
maxLeftBorderSum = leftBorderSum;
}
int maxRightBorderSum = 0, rightBorderSum = 0;
for (int i = center + 1; i <= right; i++) {
rightBorderSum += a[i];
if (rightBorderSum > maxRightBorderSum)
maxRightBorderSum = rightBorderSum;
}
return max3(maxLeftSum, maxRightSum, maxLeftBorderSum + maxRightBorderSum);
}
/**
* return the biggest of arrayList.
*/
private static int max3(int a, int b, int c) {
int max = a;
if (b > max)
max = b;
if (c > max)
max = c;
return max;
}
}