算法学习——最大连续子序列

求一组数组中和最大的连续子序列

以下展示四种方法，时间复杂度依次递减。

func1 O(N^3)

func2 O(N^2)

func3 O(NlogN)

func4 O(N)

// ConsoleApplication1.cpp: 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>

int func1(int* arr, int n) {
	int i = 0, max = 0;
	for (i = 0; i < n; i++) {
		for (int j = i; j < n; j++) {
			int sum = 0;
			for (int k = i; k <= j; k++) {
				sum += arr[k];
			}
			if (sum >= max)
				max = sum;
		}
	}
	return max;
}

int func2(int* arr, int n) {
	int i = 0, max = 0;
	for (i = 0; i < n; i++) {
		int sum = 0;
		for (int j = i; j < n; j++) {
			sum += arr[j];
			if (sum >= max)
				max = sum;
		}

	}
	return max;
}

int func3(int* arr, int left, int right) {
	int sumLeft, sumRight;
	int sumMidLeft, sumMidRight;
	int maxMidLeft, maxMidRight;
	int center;

	if (left == right) {
		if (arr[left] > 0)
			return arr[left];
		else
			return 0;
	}
	center = (left + right) / 2;
	sumLeft = func3(arr, left, center);
	sumRight = func3(arr, center + 1, right);

	sumMidLeft = 0; maxMidLeft = 0;
	for (int i = center; i >= left; i--) {
		sumMidLeft += arr[i];
		if (sumMidLeft >= maxMidLeft)
			maxMidLeft = sumMidLeft;
	}

	sumMidRight = 0; maxMidRight = 0;
	for (int i = center + 1; i <= right; i++) {
		sumMidRight += arr[i];
		if (sumMidRight >= maxMidRight)
			maxMidRight = sumMidRight;
	}
	int temp = sumLeft >= sumRight ? sumLeft : sumRight;
	return temp >= (maxMidLeft + maxMidRight) ? temp : maxMidLeft + maxMidRight;
}

int func4(int* arr, int n) {
	int i = 0, max = 0, sum = 0;
	for (i = 0; i < n; i++) {
		sum += arr[i];
		if (sum >= max)
			max = sum;
		else if(sum < 0)
			sum = 0;
	}
	return max;
}

int arr[8] = {5, 6, 7, -9, 8, -2, -10, 12};
int main()
{
	int ret = 0;
	ret = func1(arr, 8);
	printf("func1 = %d\n", ret);
	printf("func2 = %d\n", func2(arr, 8));
	printf("func3 = %d\n", func3(arr, 0, 8 - 1));
	printf("func4 = %d\n", func4(arr, 8));
	system("pause");
	return 0;
}