Maximum Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

思路一

O(N^3)时间复杂度，超时

public class Solution {
    public int maxSubArray(int[] A) {
        int maxValue = 0;
        for(int i = 0; i < A.length; i++){
            for(int j = 0; j < A.length; j++){
                int sum = 0;
                for(int k = i; k <= j; k++){                //计算sum
                    sum += A[k];
                }//for
                maxValue = maxValue > sum ? maxValue : sum;    //更新maxValue
            }//for
        }//for
        
        return maxValue;
    }
}

思路二：

O(N^2)时间复杂度，超时，ps:c++实现一样超时

public class Solution {
    public int maxSubArray(int[] A) {
        int maxValue = A[0];
        for(int i = 0; i < A.length; i++){
            int sum = A[i];
            for(int j = i + 1; j < A.length; j++){                
                sum += A[j];
                maxValue = maxValue > sum ? maxValue : sum;
            }//for
        }//for
        
        return maxValue;
    }
}

思路三：

DP，O(n)时间复杂度，可以AC

遍历数组，对于任意元素A[i]，只有两种可能，放到[k...i - 1]中，不放到[k...i - 1]中

1.放：此时sum,即连续子串的和为sum(A[k...i])

2.不放：此时sum,即连续子串的和为sum(A[i])

if(sum(A[k...i]) > sum(A[i])) 放A[i]，否则不放A[i],从A[i]开始计算sum

注意站在A[i]的角度去想

public class Solution {
    public int maxSubArray(int[] A) {
        int maxValue = A[0];
        int sum = A[0];
        for(int i = 1; i < A.length; i++){
            int temp = sum + A[i];
            sum = temp > A[i] ? temp : A[i];
            maxValue = maxValue > sum ? maxValue : sum;
        }
        
        return maxValue;
    }
}

思路四：分治法待续...

ps:相同代码，c++的时间确实要比java块很多