最大子序列和及其位置问题

/**
* 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)));
}

}