[英]Maximum Sum SubArray
我试图在具有最大总和的数组中找到连续的子数组。 所以,对于数组
{5, 15, -30, 10, -5, 40, 10}
连续使用这些数字的最大和可能是 55,或 (10 + (-5) + 40 + 10) = 55。下面的程序输出最大和 55,但是,我想弄清楚的问题是如何打印产生这个 55 的序列。换句话说,我如何打印出 10、-5、40 和 10?
public static void main(String[] args) {
int[] arr = {5, 15, -30, 10, -5, 40, 10};
System.out.println(maxSubsequenceSum(arr));
}
public static int maxSubsequenceSum(int[] X) {
int max = X[0];
int sum = X[0];
for (int i = 1; i < X.length; i++) {
sum = Math.max(X[i], sum + X[i]);
max = Math.max(max, sum);
}
return max;
}
我正在考虑创建一个 ArrayList 来存储 i 的每个索引处的sum
值,因此 ArrayList 看起来像 (5, 20, -10, 10, 5, 45, 55)。 然后我打算将 ArrayList 从索引 0 清除到列表中的第一个负数,但是,这只能解决此特定示例的问题,但是如果我更改原始数字数组,则此解决方案将不起作用。
您可以用 if 语句替换 Math.Max 函数并更新最佳子数组的开始和结束索引。 帕斯卡版本:
if X[i] > sum + X[i] then begin
sum := X[i];
start := i;
end
else
sum := sum + X[i];
if max < sum then begin
max := sum;
finish := i;
end;
您可以跟踪循环中当前最佳子数组的开始和结束索引。 不要使用max()
来计算sum
和max
,只需执行以下操作:
int sum_start = 0, sum_end = 0, start = 0, end = 0;
// In the for loop
if (X[i] > sum + X[i]) {
sum = X[i];
sum_start = i;
sum_end = i;
} else {
++sum_end;
}
if (sum > max) {
start = sum_start;
end = sum_end;
max = sum;
}
有一个 o(n) 解决方案,一个循环遍历数组并在当前总数低于 0 时重置您的子序列。
{5, 15, -30, 10, -5, 40, 10}
编辑:获取子序列...每当您更改最大值时,请更新您的子序列
可以通过在识别最大子数组的同时捕获开始和结束来完成,如下所示:
package recursion;
import java.util.Arrays;
public class MaximumSubArray {
private static SubArray maxSubArray(int[] values, int low, int high) {
if (low == high) {
// base condition
return new SubArray(low, high, values[low]);
} else {
int mid = (int) (low + high) / 2;
// Check left side
SubArray leftSubArray = maxSubArray(values, low, mid);
// Check right side
SubArray rightSubArray = maxSubArray(values, mid + 1, high);
// Check from middle
SubArray crossSubArray = maxCrossSubArray(values, low, mid, high);
// Compare left, right and middle arrays to find maximum sub-array
if (leftSubArray.getSum() >= rightSubArray.getSum()
&& leftSubArray.getSum() >= crossSubArray.getSum()) {
return leftSubArray;
} else if (rightSubArray.getSum() >= leftSubArray.getSum()
&& rightSubArray.getSum() >= crossSubArray.getSum()) {
return rightSubArray;
} else {
return crossSubArray;
}
}
}
private static SubArray maxCrossSubArray(int[] values, int low, int mid,
int high) {
int sum = 0;
int maxLeft = low;
int maxRight = high;
int leftSum = Integer.MIN_VALUE;
for (int i = mid; i >= low; i--) {
sum = sum + values[i];
if (sum > leftSum) {
leftSum = sum;
maxLeft = i;
}
}
sum = 0;
int rightSum = Integer.MIN_VALUE;
for (int j = mid + 1; j <= high; j++) {
sum = sum + values[j];
if (sum > rightSum) {
rightSum = sum;
maxRight = j;
}
}
SubArray max = new SubArray(maxLeft, maxRight, (leftSum + rightSum));
return max;
}
static class SubArray {
private int start;
private int end;
private int sum;
public SubArray(int start, int end, int sum) {
super();
this.start = start;
this.end = end;
this.sum = sum;
}
public int getStart() { return start; }
public void setStart(int start) { this.start = start; }
public int getEnd() { return end; }
public void setEnd(int end) { this.end = end; }
public int getSum() { return sum; }
public void setSum(int sum) { this.sum = sum; }
@Override
public String toString() {
return "SubArray [start=" + start + ", end=" + end + ", sum=" + sum + "]";
}
}
public static final void main(String[] args) {
int[] values = { 5, 15, -30, 10, -5, 40, 10 };
System.out.println("Maximum sub-array for array"
+ Arrays.toString(values) + ": " + maxSubArray(values, 0, 6));
}
}
Maximum sub-array for array[5, 15, -30, 10, -5, 40, 10]: SubArray [start=3, end=6, sum=55]
解决方案可以从https://github.com/gosaliajigar/Programs/blob/master/src/recursion/MaximumSubArray.java下载
两个子阵列是
[1, 2, 3]
和 [4, 9] 不包括负数
这里的最大子数组是 [ 4, 5]
所以输出是9
这是代码
public class MaxSubArray{
static void sumM(int a[], int n){
int s1 = Integer.MAX_VALUE;
int k = Integer.MAX_VALUE;
int sum = 0;
int s2 = 0;
for(int i=0;i<n;i++){
if(a[i]<s1){
if(a[i]<0){
k = Math.min(a[i],s1);
}
}
if(a[i]>k){
sum+=a[i];
}
if(a[i]<k){
if(a[i]<0){
continue;
}
s2+=a[i];
}
}
if(sum>s2){
System.out.println(sum);
}
else{
System.out.println(s2);
}
}
public static void main(String[] args){
int a[] = {1,2,3,-7,4,5};
int n = a.length;
sumM(a,n);
}
}
public static int kadane(int[] A) {
int maxSoFar = 0;
int maxEndingHere = 0;
// traverse the given array
for (int i: A) {
// update the maximum sum of subarray "ending" at index `i` (by adding the
// current element to maximum sum ending at previous index `i-1`)
maxEndingHere = maxEndingHere + i;
// if the maximum sum is negative, set it to 0 (which represents
// an empty subarray)
maxEndingHere = Integer.max(maxEndingHere, 0);
// update the result if the current subarray sum is found to be greater
maxSoFar = Integer.max(maxSoFar, maxEndingHere);
}
return maxSoFar;
}
您需要对所有可能的子数组求和。 要做到这一点,你可以做这个代码
public static int maxSubsequenceSum(int[] X) {
int max = 0;
boolean max_init = false;
int max_from=0;
int max_to=0; // this is not included
for (int i = 0; i < X.length; i++) {
for (int j = i + 1; j < X.length + 1; j++) {
int total = 0;
for (int k = i; k < j; k++) {
total += X[k];
}
if (total > max || !max_init){
max = total;
max_init = true;
max_from = i;
max_to = j;
}
}
}
for (int i=max_from;i<max_to;i++)
System.out.print(X[i]+",");
System.out.println();
return max;
}
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