我想用展示（n log n）的算法实现最大子阵列问题：
找出最大的连续子数组，或数组中连续元素的最大总和。
假设：并非所有元素都是负数
我有点工作的解决方案;问题在于重叠的中心数组，以及指定重叠子问题的适当索引，一些数组我得到正确答案而不是其他问题。
仅供比较＆amp;为了获得正确性，我实施了一种称为Kadane算法的解决方案 （我相信复杂性是Omega（n））。
这是Kandane的算法（http://en.wikipedia.org/wiki/Maximum_subarray_problem）：
public static void Kadane(int array[]) {
int max_ending_here = 0;
for (int i = 0; i < array.length; i++) {
max_ending_here = max_ending_here + array[i];
max_so_far = Math.max(max_so_far, max_ending_here);
}
System.out.println("Kadane(int array []): " + max_so_far);
}
我的递归实现，使用divide＆amp;征服比较子数组的最大值，然后对具有最大值的子数组进行递归调用，直到递归结束
public static void findMaxSubArray(int[] array, int lowIndex, int highIndex) {
int mid = 0;
int arrayLength = 0;
int maxEndingHere = 0;
if (array == null) {
throw new NullPointerException();
} else if
//base condition
(array.length < 2 || (highIndex==lowIndex)) {
maxLowIndex = lowIndex;
maxHighIndex = highIndex;
System.out.println("findMaxSubArray(int[] array, int lowIndex, int highIndex)");
System.out.println("global Max Range, low:" + maxLowIndex + " high: " + maxHighIndex);
System.out.println("global Max Sum:" + globalMaximum);
} else {
System.out.println();
int lowMidMax = 0;
int midHighMax = 0;
int centerMax = 0;
//array length is always the highest index +1
arrayLength = highIndex + 1;
//if even number of elements in array
if (array.length % 2 == 0) {
mid = arrayLength / 2;
System.out.println("low: " + lowIndex + " mid: " + mid);
for (int i = lowIndex; i < mid; i++) {
System.out.print(array[i] + ",");
}
//calculate maximum contigous array encountered so far in low to mid indexes
for (int i = lowIndex; i < mid; i++) {
maxEndingHere = maxEndingHere + array[i];
if (maxEndingHere < 0) {
maxEndingHere = 0;
}
if (globalMaximum < maxEndingHere) {
//new maximum found
globalMaximum = lowMidMax = maxEndingHere;
lowIndex = i;
}
}
//end low mid calc
for (int i = mid; i <= highIndex; i++) {
System.out.print(array[i] + ",");
}
System.out.println("mid: " + mid + " high: " + highIndex);
//calculate maximum contigous array encountered so far in mid to high indexes
for (int i = mid; i <= highIndex; i++) {
maxEndingHere = maxEndingHere + array[i];
if (maxEndingHere < 0) {
maxEndingHere = 0;
}
if (globalMaximum < maxEndingHere) {
//new maximum found
globalMaximum = midHighMax = maxEndingHere;
mid = i;
}
}
//end mid high calc
//calculate maximum contigous array encountered so far in center array
int lowCenter = mid -1;
int highCenter = highIndex -1;
System.out.println("lowCenter: " + lowCenter + " highCenter: " + highCenter);
for (int i = lowCenter; i < highCenter; i++) {
System.out.print(array[i] + ",");
}
//calculate maximum contigous array encountered so far in low to mid indexes
for (int i = lowCenter; i < highCenter; i++) {
maxEndingHere = maxEndingHere + array[i];
if (maxEndingHere < 0) {
maxEndingHere = 0;
}
if (globalMaximum < maxEndingHere) {
//new max found
globalMaximum = centerMax = maxEndingHere;
lowCenter = i;
}
}
//end center calc
//determine which range contains the maximum sub array
//if lowMidMax <= midHighMax and centerMax
if (lowMidMax >= midHighMax && lowMidMax >= centerMax) {
maxLowIndex = lowIndex;
maxHighIndex = mid;
//recursive call
findMaxSubArray(array, lowIndex, mid);
}
if (midHighMax >= lowMidMax && midHighMax >= centerMax) {
maxLowIndex = mid;
maxHighIndex = highIndex;
//recursive call
findMaxSubArray(array, mid, highIndex);
}
if (centerMax >= midHighMax && centerMax >= lowMidMax) {
maxLowIndex = lowCenter;
maxHighIndex = highCenter;
//recursive call
findMaxSubArray(array, lowCenter, highCenter);
}
}//end if even parent array
//else if uneven array
else {
mid = (int) Math.floor(arrayLength / 2);
System.out.println("low: " + lowIndex + " mid: " + mid);
for (int i = lowIndex; i < mid; i++) {
System.out.print(array[i] + ",");
}
//calculate maximum contigous array encountered so far in low to mid indexes
for (int i = lowIndex; i < mid; i++) {
maxEndingHere = maxEndingHere + array[i];
if (maxEndingHere < 0) {
maxEndingHere = 0;
}
if (globalMaximum < maxEndingHere) {
//new maximum found
globalMaximum = lowMidMax = maxEndingHere;
lowIndex = i;
}
}
//end low mid calc
System.out.println("mid+1: " + (mid + 1) + " high: " + highIndex);
for (int i = mid + 1; i <= highIndex; i++) {
System.out.print(array[i] + ",");
}
//calculate maximum contigous array encountered so far in mid to high indexes
for (int i = mid + 1; i <= highIndex; i++) {
maxEndingHere = maxEndingHere + array[i];
if (maxEndingHere < 0) {
maxEndingHere = 0;
}
if (globalMaximum < maxEndingHere) {
//new maximum found
globalMaximum = midHighMax = maxEndingHere;
mid = i;
}
}
//end mid high calc
//calculate maximum contigous array encountered so far in center array
int lowCenter = mid;
int highCenter = highIndex -1;
System.out.println("lowCenter: " + lowCenter + " highCenter: " + highCenter);
for (int i = lowCenter; i < highCenter; i++) {
System.out.print(array[i] + ",");
}
//calculate maximum contigous array encountered so far in low to mid indexes
for (int i = lowCenter; i < highCenter; i++) {
maxEndingHere = maxEndingHere + array[i];
if (maxEndingHere < 0) {
maxEndingHere = 0;
}
if (globalMaximum < maxEndingHere) {
//new max
globalMaximum = centerMax = maxEndingHere;
lowCenter = i;
}
}
//end center calc
//determine which range contains the maximum sub array
//if lowMidMax <= midHighMax and centerMax
if (lowMidMax >= midHighMax && lowMidMax >= centerMax) {
maxLowIndex = lowIndex;
maxHighIndex = mid;
//recursive call
findMaxSubArray(array, lowIndex, mid);
}
if (midHighMax >= lowMidMax && midHighMax >= centerMax) {
maxLowIndex = mid;
maxHighIndex = highIndex;
//recursive call
findMaxSubArray(array, mid, highIndex);
}
if (centerMax >= midHighMax && centerMax >= lowMidMax) {
maxLowIndex = lowCenter;
maxHighIndex = highCenter;
//recursive call
findMaxSubArray(array, lowCenter, highCenter);
}
}//end odd parent array length
}//end outer else array is ok to computed
}//end method
结果：使用数组subArrayProblem1 = {1,2,3,4,5,6,7,8};
Kadane（int array []）：36 低：0中：4 1,2,3,4,5,6,7,8，中：4高：7 lowCenter：6 highCenter：6
findMaxSubArray（int [] array，int lowIndex，int highIndex） 全局最大范围，低：7高：7 全球最高金额：36 建立成功（总时间：0秒）
问题虽然与Kadane相比，全局最大总和是正确的，但是低指数＆amp;高指数范围反映了最后一次重复调用。
结果：使用数组subArrayProblem = {100,113,110,85,105,102,86,63,81,101,94,106,101,79,94,90,97};
Kadane（int array []）：1607 低：0中：8 100,113,110,85,105,102,86,63，中+ 1：9高：16 101,94,106,101,79,94,90,97，lowCenter：16 highCenter：15
findMaxSubArray（int [] array，int lowIndex，int highIndex） 全局最大范围，低：16高：16 全球最大总和：1526
全局最大值不正确，注意差异实际上是1个元素，即元素81
答案 0 :(得分：1)
1. Kadane算法的实现是错误的，它会在带有一些负数的数组上失败。 正确的应该是这样的：
public static void Kadane(int array[]) {
int max_ending_here = 0;
for (int i = 0; i < array.length; i++) {
max_ending_here = Math.max(array[i], max_ending_here + array[i]);
max_so_far = Math.max(max_so_far, max_ending_here);
}
System.out.println("Kadane(int array []): " + max_so_far);
}
您的代码中存在许多错误，例如：
2.在计算答案之前，maxEndingHere应初始化为0：
[lowIndex,mid)
[mid, highIndex]
[lowCenter, highCenter]
现在它只在第一次迭代之前被初始化。
3. lowCenter应初始化为lowIndex
4.该程序不必要地冗长而复杂......我不知道我是否错过了任何错误......
答案 1 :(得分：1)
解决方案非常简单，pv是一个变量，它向我们展示了从k开始的子数组的总和，其中0 <= k <= n-1到我们在第i次迭代中访问的变量。数组s让我们跟踪最大子数组，s [i]也是在第i次迭代中找到的最大子数组。 c是原始数组。 p是指向在第i次迭代中找到的最大子数组的开始的指针。 tp是将p更新为正确值的临时指针
/**
* @author : Yash M. Sawant
*/
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define MAXLENGTH 17
#define MINVALUE -999
int main() {
int i;
int s[MAXLENGTH]; s[0] = 0;
int c[MAXLENGTH] = {0, 13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7};
int pv = 0, p = -1, tp = -1;
for(i = 1 ; i < MAXLENGTH ; i ++) {
printf("%4d ", c[i]);
if(s[i - 1] < pv + c[i]) {
s[i] = pv + c[i];
pv = pv + c[i];
if(tp > p) {
p = tp;
}
} else {
s[i] = s[i - 1];
pv = pv + c[i];
if(pv < 0) {
pv = 0; tp = i;
}
}
}
printf("\n");
for(i = 0 ; i < MAXLENGTH ; i ++) {
printf("%4d ", s[i]);
}
printf("\n");
printf("Max Sub Array = %d and Starts at %d ", s[MAXLENGTH - 1], p);
return 0;
}
答案 2 :(得分：1)
更干净的方法来找到具有最大和的子数组（D / C递归方法）：
MATCH()