原文:
给定一个数组arr【】，任务是找到具有最大和的相邻数字子数组的元素。
示例:
输入: arr = [-2，-3，4，-1，-2，1，5，-3] 输出:【4，-1，-2，1，5】 解释: 在上述输入中，最大连续子阵列和为 7，子阵列的元素为【4，-1，-2，1，5】
输入: arr = [-2，-5，6，-2，-3，1，5，-6] 输出:【6，-2，-3，1，5】 解释: 在上述输入中，最大连续子阵列和为 7，子阵列的元素 为【6，-2，-3，1，5】
天真方法:天真方法是生成所有可能的子阵列并打印具有最大和的那个子阵列。 时间复杂度:o(n2) 辅助空间: o(1)
高效途径:思路是利用找到最大子阵和，并存储具有最大和的子阵的起始和结束索引，打印从起始索引到结束索引的子阵。以下是步骤:
- 将 3 个变量 endindex 初始化为 0， currmax 和 globalmax 初始化为输入数组的第一个值。
- 对于数组中从索引(比如 i ) 1 开始的每个元素，更新 currmax 到 max(nums[i]，nums[i] currmax) 和 globalmax 和 endindex 到 i 仅当 currmax > globalmax 时。
- 要找到起始索引，从 endindex 开始向左迭代，并不断递减 globalmax 的值，直到它变成 0。它变为 0 的点是开始索引。
- 现在打印【开始，结束】之间的子阵列。
下面是上述方法的实现:
c
// c program for the above approach
#include
using namespace std;
// function to print the elements
// of subarray with maximum sum
void subarraywithmaxsum(vector& nums)
{
// initialize currmax and globalmax
// with first value of nums
int endindex, currmax = nums[0];
int globalmax = nums[0];
// iterate for all the elements
// of the array
for (int i = 1; i < nums.size(); i) {
// update currmax
currmax = max(nums[i],
nums[i] currmax);
// check if currmax is greater
// than globalmax
if (currmax > globalmax) {
globalmax = currmax;
endindex = i;
}
}
int startindex = endindex;
// traverse in left direction to
// find start index of subarray
while (startindex >= 0) {
globalmax -= nums[startindex];
if (globalmax == 0)
break;
// decrement the start index
startindex--;
}
// printing the elements of
// subarray with max sum
for (int i = startindex;
i <= endindex; i) {
cout << nums[i] << " ";
}
}
// driver code
int main()
{
// given array arr[]
vector arr
= { -2, -5, 6, -2,
-3, 1, 5, -6 };
// function call
subarraywithmaxsum(arr);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program for the above approach
import java.util.*;
class gfg{
// function to print the elements
// of subarray with maximum sum
static void subarraywithmaxsum(vector nums)
{
// initialize currmax and globalmax
// with first value of nums
int endindex = 0, currmax = nums.get(0);
int globalmax = nums.get(0);
// iterate for all the elements
// of the array
for (int i = 1; i < nums.size(); i)
{
// update currmax
currmax = math.max(nums.get(i),
nums.get(i) currmax);
// check if currmax is greater
// than globalmax
if (currmax > globalmax)
{
globalmax = currmax;
endindex = i;
}
}
int startindex = endindex;
// traverse in left direction to
// find start index of subarray
while (startindex >= 0)
{
globalmax -= nums.get(startindex);
if (globalmax == 0)
break;
// decrement the start index
startindex--;
}
// printing the elements of
// subarray with max sum
for(int i = startindex; i <= endindex; i)
{
system.out.print(nums.get(i) " ");
}
}
// driver code
public static void main(string[] args)
{
// given array arr[]
vector arr = new vector();
arr.add(-2);
arr.add(-5);
arr.add(6);
arr.add(-2);
arr.add(-3);
arr.add(1);
arr.add(5);
arr.add(-6);
// function call
subarraywithmaxsum(arr);
}
}
// this code is contributed by rajput-ji
python 3
# python3 program for the above approach
# function to print the elements
# of subarray with maximum sum
def subarraywithmaxsum(nums):
# initialize currmax and globalmax
# with first value of nums
currmax = nums[0]
globalmax = nums[0]
# iterate for all the elements
# of the array
for i in range(1, len(nums)):
# update currmax
currmax = max(nums[i],
nums[i] currmax)
# check if currmax is greater
# than globalmax
if (currmax > globalmax):
globalmax = currmax
endindex = i
startindex = endindex
# traverse in left direction to
# find start index of subarray
while (startindex >= 0):
globalmax -= nums[startindex]
if (globalmax == 0):
break
# decrement the start index
startindex -= 1
# printing the elements of
# subarray with max sum
for i in range(startindex, endindex 1):
print(nums[i], end = " ")
# driver code
# given array arr[]
arr = [ -2, -5, 6, -2,
-3, 1, 5, -6 ]
# function call
subarraywithmaxsum(arr)
# this code is contributed by sanjoy_62
c
// c# program for the above approach
using system;
using system.collections.generic;
class gfg{
// function to print the elements
// of subarray with maximum sum
static void subarraywithmaxsum(list nums)
{
// initialize currmax and globalmax
// with first value of nums
int endindex = 0, currmax = nums[0];
int globalmax = nums[0];
// iterate for all the elements
// of the array
for (int i = 1; i < nums.count; i)
{
// update currmax
currmax = math.max(nums[i],
nums[i] currmax);
// check if currmax is greater
// than globalmax
if (currmax > globalmax)
{
globalmax = currmax;
endindex = i;
}
}
int startindex = endindex;
// traverse in left direction to
// find start index of subarray
while (startindex >= 0)
{
globalmax -= nums[startindex];
if (globalmax == 0)
break;
// decrement the start index
startindex--;
}
// printing the elements of
// subarray with max sum
for(int i = startindex; i <= endindex; i)
{
console.write(nums[i] " ");
}
}
// driver code
public static void main(string[] args)
{
// given array []arr
list arr = new list();
arr.add(-2);
arr.add(-5);
arr.add(6);
arr.add(-2);
arr.add(-3);
arr.add(1);
arr.add(5);
arr.add(-6);
// function call
subarraywithmaxsum(arr);
}
}
// this code is contributed by gauravrajput1
java 描述语言
output
6 -2 -3 1 5
时间复杂度: o(n) 辅助空间: o(1)
交替高效方法:该方法消除了上述方法中向左递减 global_max 的内部 while 循环。因此这种方法减少了时间。
- 将 currmax 和 globalmax 初始化为输入数组的第一个值。将 endindex、startindex、globalmaxstartindex 初始化为 0。(endindex，startindex 存储以 i 结尾的最大和子数组的开始和结束索引。globalmaxstartindex 存储 globalmax 的开始索引)
- 对于数组中从 index(比如 i) 1 开始的每个元素，如果 nums[i] > nums[i] currmax，则将 currmax 和 startindex 更新为 i。否则只更新 currmax。
- 如果当前最大值>全局最大值，则更新全局最大值。同时更新 globalmaxstartindex。
- 现在打印[开始索引，全局最开始索引]之间的子数组。
以下是上述方法的实现:
java 语言(一种计算机语言，尤用于创建网站)
// java program for the above approach
import java.util.*;
class gfg {
// function to print the elements
// of subarray with maximum sum
static void subarraywithmaxsum(vector nums)
{
// initialize currmax and globalmax
// with first value of nums
int currmax = nums.get(0), globalmax = nums.get(0);
// initialize endindex startindex,globalstartindex
int endindex = 0;
int startindex = 0, globalmaxstartindex = 0;
// iterate for all the elements of the array
for (int i = 1; i < nums.size(); i) {
// update currmax and startindex
if (nums.get(i) > nums.get(i) currmax) {
currmax = nums.get(i);
startindex = i; // update the new startindex
}
// update currmax
else if (nums.get(i) < nums.get(i) currmax) {
currmax = nums.get(i) currmax;
}
// update globalmax anf globalmaxstartindex
if (currmax > globalmax) {
globalmax = currmax;
endindex = i;
globalmaxstartindex = startindex;
}
}
// printing the elements of subarray with max sum
for (int i = globalmaxstartindex; i <= endindex;
i) {
system.out.print(nums.get(i) " ");
}
}
// driver code
public static void main(string[] args)
{
// given array arr[]
vector arr = new vector();
arr.add(-2);
arr.add(-5);
arr.add(6);
arr.add(-2);
arr.add(-3);
arr.add(1);
arr.add(5);
arr.add(-6);
// function call
subarraywithmaxsum(arr);
}
}
// this code is contributed by amritha basavaraj patil
c
// c# program for the above approach
using system;
using system.collections.generic;
public class gfg{
// function to print the elements
// of subarray with maximum sum
static void subarraywithmaxsum(list nums)
{
// initialize currmax and globalmax
// with first value of nums
int currmax = nums[0], globalmax = nums[0];
// initialize endindex startindex,globalstartindex
int endindex = 0;
int startindex = 0, globalmaxstartindex = 0;
// iterate for all the elements of the array
for (int i = 1; i < nums.count; i) {
// update currmax and startindex
if (nums[i] > nums[i] currmax) {
currmax = nums[i];
startindex = i; // update the new startindex
}
// update currmax
else if (nums[i] < nums[i] currmax) {
currmax = nums[i] currmax;
}
// update globalmax anf globalmaxstartindex
if (currmax > globalmax) {
globalmax = currmax;
endindex = i;
globalmaxstartindex = startindex;
}
}
// printing the elements of subarray with max sum
for (int i = globalmaxstartindex; i <= endindex;
i) {
console.write(nums[i] " ");
}
}
// driver code
static public void main (){
// given array arr[]
list arr = new list();
arr.add(-2);
arr.add(-5);
arr.add(6);
arr.add(-2);
arr.add(-3);
arr.add(1);
arr.add(5);
arr.add(-6);
// function call
subarraywithmaxsum(arr);
}
}
// this code is contributed by rag2127.
java 描述语言
output
6 -2 -3 1 5
时间复杂度:*o(n)t5】 t8】辅助空间: o(1)***
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