lintcode-41. Maximum Subarray

1. 问题描述

Description

Given an array of integers, find a contiguous subarray which has the largest sum.

The subarray should contain at least one number.

2. my solution

2.1 我的思路

因为是连续的, 所以想到位置不断向前移动,和小于0更新位置, 记录最大和的方法

2.2 代码实现

public class Solution {
    /**
     * @param nums: A list of integers
     * @return: A integer indicate the sum of max subarray
     */
    public int maxSubArray(int[] nums) {
        // write your code here
        int sum = 0;
        int maxSum = nums[0];
        
        for (int i = 0;i<nums.length ;i++ ) 
        {
            sum += nums[i];
            if(sum>maxSum)
                maxSum = sum;
            else if(sum<=0) sum = 0;
        }
        return maxSum;
    }
}

2.3 运行结果

3. others solutions

3.1 思路一(较好)

public class Solution {
    public int maxSubArray(int[] A) {
        if (A == null || A.length == 0){
            return 0;
        }
        //max记录全局最大值，sum记录前i个数的和，minSum记录前i个数中0-k的最小值
        int max = Integer.MIN_VALUE, sum = 0, minSum = 0;
        for (int i = 0; i < A.length; i++) {
            sum += A[i];
            max = Math.max(max, sum - minSum);
            minSum = Math.min(minSum, sum);
        }

        return max;
    }
}