Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example:

Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.

Follow up:

        If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

方法1：這種方式比較好理解，也是我認為效率最高的演算法

package leetcode;

import org.junit.Test;

/**
 * @author zhangyu
 * @version V1.0
 * @ClassName: MaximumSubarray
 * @Description:
 * @date 2018/12/6 9:41
 **/


public class MaximumSubarray2 {
    @Test
    public void fun() {
        int[] nums = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
        int maxNum = maximumSubarray(nums);
        System.out.println(maxNum);
    }

    //這種比較直接，又通俗易懂的方式
    private int maximumSubarray(int[] nums) {
        int max = nums[0], sum = 0;
        for (int i = 0; i < nums.length; i++) {
            if (sum < 0) {
                sum = nums[i];
            } else {
                sum += nums[i];
            }
            if (sum > max) {
                max = sum;
            }
        }
        return max;
    }
}
 

時間複雜度：O(n)

空間複雜度：O(1)

方法2：利用動態規劃，最主要是狀態轉移方程

package leetcode;

import org.junit.Test;

/**
 * @author zhangyu
 * @version V1.0
 * @ClassName: MaximumSubarray
 * @Description: 這種是採用動態規劃的方式進行求解
 * @date 2018/12/6 9:41
 **/


public class MaximumSubarray3 {
    @Test
    public void fun() {
        int[] nums = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
        int maxNum = maximumSubarray(nums);
        System.out.println(maxNum);
    }

    private int maximumSubarray(int[] nums) {
        // 如果陣列為空或者長度為0，直接返回0
        if (nums == null || nums.length == 0) {
            return 0;
        }
        // 定義最大值為陣列開始的數
        int result = nums[0];
        int sum = nums[0];
        for (int i = 1; i < nums.length; ++i) {
            // 主要是要想明白這個動態規劃的狀態轉換方程
            sum = Math.max(sum + nums[i], nums[i]);
            result = Math.max(result, sum);
        }
        // 最後返回最大的那個數
        return result;
    }
}
 

時間複雜度：O(n)

空間複雜度：O(n)