常量定义
private static int seqStart = 0;
private static int seqEnd = -1;
O(N3)算法
注意seqStart和seqEnd代表实际的最佳顺序。
public static int maxSubSum1(int[] a) {
int maxSum = 0;
for(int i = 0; i < a.length; i++) {
for(int j = i; j < a.length; j++) {
int thisSum = 0;
for(int k = i; k <= j; k++) {
thisSum += a[k];
}
if(thisSum > maxSum) {
maxSum = thisSum;
seqStart = i;
seqEnd = j;
}
}
}
return maxSum;
}
O(N2)算法
注意seqStart和seqEnd代表实际的最佳顺序。
public static int maxSubSum2(int[] a) {
int maxSum = 0;
for(int i = 0; i < a.length; i++) {
int thisSum = 0;
for(int j = i; j < a.length; j++) {
thisSum += a[j];
if(thisSum > maxSum) {
maxSum = thisSum;
seqStart = i;
seqEnd = j;
}
}
}
return maxSum;
}
O(N)算法
注意seqStart和seqEnd代表实际的最佳顺序。
public static int maxSubSum4(int[] a) {
int maxSum = 0;
int thisSum = 0;
for(int i = 0, j = 0; j < a.length; j++) {
thisSum += a[j];
if(thisSum > maxSum) {
maxSum = thisSum;
seqStart = i;
seqEnd = j;
} else if(thisSum < 0) {
i = j + 1;
thisSum = 0;
}
}
return maxSum;
}
递归版本
在跨越a [left…right]的子数组中找到最大和。
不尝试维持实际的最佳顺序。
调用递归的函数：
public static int maxSubSum3(int[] a) {
return a.length > 0 ? maxSumRec( a, 0, a.length - 1 ) : 0;
}
递归部分：
private static int maxSumRec(int[] a, int left, int right) {
int maxLeftBorderSum = 0, maxRightBorderSum = 0;
int leftBorderSum = 0, rightBorderSum = 0;
int center = left + right >>> 1;
if(left == right) {
return a[left] > 0 ? a[left] : 0;
}
int maxLeftSum = maxSumRec(a, left, center);
int maxRightSum = maxSumRec(a, center + 1, right);
for(int i = center; i >= left; i--) {
leftBorderSum += a[i];
if(leftBorderSum > maxLeftBorderSum) {
maxLeftBorderSum = leftBorderSum;
}
}
for(int i = center + 1; i <= right; i++) {
rightBorderSum += a[i];
if(rightBorderSum > maxRightBorderSum) {
maxRightBorderSum = rightBorderSum;
}
}
return max3(maxLeftSum, maxRightSum, maxLeftBorderSum + maxRightBorderSum);
}
辅助函数——三数求Max：
private static int max3(int a, int b, int c) {
return a > b ? a > c ? a : c : b > c ? b : c;
}
完整代码
import java.util.Random;
public final class MaxSumTest {
private static int seqStart = 0;
private static int seqEnd = -1;
/**
* Cubic maximum contiguous subsequence sum algorithm.
* seqStart and seqEnd represent the actual best sequence.
*/
public static int maxSubSum1(int[] a) {
int maxSum = 0;
for(int i = 0; i < a.length; i++) {
for(int j = i; j < a.length; j++) {
int thisSum = 0;
for(int k = i; k <= j; k++) {
thisSum += a[k];
}
if(thisSum > maxSum) {
maxSum = thisSum;
seqStart = i;
seqEnd = j;
}
}
}
return maxSum;
}
/**
* Quadratic maximum contiguous subsequence sum algorithm.
* seqStart and seqEnd represent the actual best sequence.
*/
public static int maxSubSum2(int[] a) {
int maxSum = 0;
for(int i = 0; i < a.length; i++) {
int thisSum = 0;
for(int j = i; j < a.length; j++) {
thisSum += a[j];
if(thisSum > maxSum) {
maxSum = thisSum;
seqStart = i;
seqEnd = j;
}
}
}
return maxSum;
}
/**
* Linear-time maximum contiguous subsequence sum algorithm.
* seqStart and seqEnd represent the actual best sequence.
*/
public static int maxSubSum4(int[] a) {
int maxSum = 0;
int thisSum = 0;
for(int i = 0, j = 0; j < a.length; j++) {
thisSum += a[j];
if(thisSum > maxSum) {
maxSum = thisSum;
seqStart = i;
seqEnd = j;
} else if(thisSum < 0) {
i = j + 1;
thisSum = 0;
}
}
return maxSum;
}
/**
* 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) {
int maxLeftBorderSum = 0, maxRightBorderSum = 0;
int leftBorderSum = 0, rightBorderSum = 0;
int center = left + right >>> 1;
if(left == right) {
return a[left] > 0 ? a[left] : 0;
}
int maxLeftSum = maxSumRec(a, left, center);
int maxRightSum = maxSumRec(a, center + 1, right);
for(int i = center; i >= left; i--) {
leftBorderSum += a[i];
if(leftBorderSum > maxLeftBorderSum) {
maxLeftBorderSum = leftBorderSum;
}
}
for(int i = center + 1; i <= right; i++) {
rightBorderSum += a[i];
if(rightBorderSum > maxRightBorderSum) {
maxRightBorderSum = rightBorderSum;
}
}
return max3(maxLeftSum, maxRightSum, maxLeftBorderSum + maxRightBorderSum);
}
/**
* Return maximum of three integers.
*/
private static int max3(int a, int b, int c) {
return a > b ? a > c ? a : c : b > c ? b : c;
}
/**
* Driver for divide-and-conquer maximum contiguous
* subsequence sum algorithm.
*/
public static int maxSubSum3(int[] a) {
return a.length > 0 ? maxSumRec( a, 0, a.length - 1 ) : 0;
}
public static void getTimingInfo(int n, int alg) {
int [] test = new int[n];
long startTime = System.currentTimeMillis();;
long totalTime = 0;
int i;
for(i = 0; totalTime < 4000; i++) {
for(int j = 0; j < test.length; j++) {
test[j] = rand.nextInt(100) - 50;
}
switch(alg) {
case 1:
maxSubSum1(test);
break;
case 2:
maxSubSum2(test);
break;
case 3:
maxSubSum3(test);
break;
case 4:
maxSubSum4(test);
break;
}
totalTime = System.currentTimeMillis() - startTime;
}
System.out.print(String.format("\t%12.6f", (totalTime * 1000 / i) / (double)1000000));
}
private static Random rand = new Random();
/**
* Simple test program.
*/
public static void main(String[] args) {
int a[] = { 4, -3, 5, -2, -1, 2, 6, -2 };
int maxSum;
maxSum = maxSubSum1(a);
System.out.println("Max sum is " + maxSum + "; it goes" + " from " + seqStart + " to " + seqEnd);
maxSum = maxSubSum2(a);
System.out.println("Max sum is " + maxSum + "; it goes" + " from " + seqStart + " to " + seqEnd);
maxSum = maxSubSum3(a);
System.out.println("Max sum is " + maxSum);
maxSum = maxSubSum4(a);
System.out.println("Max sum is " + maxSum + "; it goes" + " from " + seqStart + " to " + seqEnd);
for(int n = 100; n <= 1000000; n *= 10) {
System.out.print(String.format("N = %7d" , n));
for(int alg = 1; alg <= 4; alg++) {
if(alg == 1 && n > 50000) {
System.out.print("\t NA ");
continue;
}
getTimingInfo(n, alg);
}
System.out.println();
}
}
}
测试结果
Max sum is 11; it goes from 0 to 6
Max sum is 11; it goes from 0 to 6
Max sum is 11
Max sum is 11; it goes from 0 to 6
N = 100 0.000040 0.000002 0.000002 0.000001
N = 1000 0.041708 0.000149 0.000025 0.000010
N = 10000 43.274000 0.019925 0.000353 0.000108
N = 100000 NA 1.428666 0.002906 0.001066
N = 1000000 NA 147.347000 0.032411 0.011284
