最大子序列和及其位置问题

最大子序列和及其位置

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/**
 * 文件名：MaxSubSumText.java
 * 时间：2014年10月21日下午6:09:36
 * 作者：修维康
 */
package chapter3;
import java.util.*;
/**
 * 类名：MaxSubSumTest.java 
 * 说明：求最大子序列和的几种算法
 */
public class MaxSubSumTest {
	/**
	 * 函数名称：maxSubSum1
	 *  说明：时间复杂度O(n^3)
	 */
	static int[] maxSubSum1(int[] a) {
		int max = 0;// 最小的返回0
		int[] result = null;// result[0]代表序列的开始下标[1]代表结束下标[2]代表和
		for (int i = 0; i < a.length; i++) {
			for (int j = i; j < a.length; j++) {
				int sum = 0;
				for (int k = i; k <= j; k++) {
					sum += a[k];
				}
				if (sum > max) {
					max = sum;
					result = new int[] { i, j, max };
				}
			}
		}
		return result;
	}
	/**
	 * 函数名称：maxSubSum2 
	 * 说明：时间复杂度O(n^2)
	 */
	static int[] maxSubSum2(int[] a) {
		int max = 0;
		int[] result = null;
		for (int i = 0; i < a.length; i++) {
			int sum = 0;
			for (int j = i; j < a.length; j++) {
				sum = sum + a[j];
				if (sum > max) {
					max = sum;
					result = new int[] { i, j, max };
				}
			}
		}
		return result;
	}
	/**
	 * 函数名称：maxSubSum3 
	 * 说明：递归分治求解 时间复杂度O(NlogN)
	 */
	static int[] maxSubSum3(int[] a, int left, int right) {
		if (left == right)// 最基础的（递归返回条件）
			return new int[] { left, right, a[left] };
		int center = (left + right) / 2;
		int[] maxLeft = maxSubSum3(a, left, center);
		int[] maxRight = maxSubSum3(a, center + 1, right);
		int maxLeftBorderSum = -1000;
		int leftBorderSum = 0;
		int leftFlag = 0;// 最大的左边位置
		for (int i = center; i >= left; i--) {
			leftBorderSum += a[i];
			if (leftBorderSum > maxLeftBorderSum) {
				maxLeftBorderSum = leftBorderSum;
				leftFlag = i;
			}
		}
		int[] result = new int[3];// result数组记录跨过中间的
		result[0] = leftFlag;
		int maxRightBorderSum = -1000;
		int rightBorderSum = 0;
		int rightFlag = 0;
		for (int i = center + 1; i <= right; i++) {
			rightBorderSum += a[i];
			if (rightBorderSum > maxRightBorderSum) {
				maxRightBorderSum = rightBorderSum;
				rightFlag = i;
			}
		}
		result[1] = rightFlag;
		result[2] = maxRightBorderSum + maxLeftBorderSum;
		return (maxLeft[2] > maxRight[2] ? maxLeft : maxRight)[2] > result[2] ? (maxLeft[2] > maxRight[2] ? maxLeft
				: maxRight)
				: result;
	}
	/**
	 * 函数名称：maxSubSum4 
	 * 说明：时间复杂度O(N)//
	 */
	public static int[] maxSubSum4(int a[]) {
		int[] result = new int[3];
		int sum = 0, flag = 0;
		result[2] = 0;
		for (int i = 0; i < a.length; i++) {
			sum += a[i];
			if (sum > result[2]) {
				result[2] = sum;
				if ((flag++) == 0)
					result[0] = i;
				result[1] = i;
			} else if (sum < 0) {
				sum = 0;
				flag = 0;
			}
		}
		return result;
	}
	public static void main(String[] args) {
		// TODO 自动生成的方法存根
		int[] a = new int[] { -1, 2, -5, 7, 8, -5 };
		System.out.println(Arrays.toString(maxSubSum1(a)));
		System.out.println(Arrays.toString(maxSubSum4(a)));
		System.out.println(Arrays.toString(maxSubSum3(a, 0, a.length - 1)));
		System.out.println(Arrays.toString(maxSubSum4(a)));
	}
}