LeetCode Maximum Subarray

LeetCode 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.

思路分析：还是考察DP，定义数组sum[i]保存从A[0]到A[i]的最大连续子数组sum

错误的解法：

sum[i] = Math.max(sum[i-1], sum[i-1] + A[i])

比如针对[-1,2]会输出-1,但是正确结果应该是输出2. 所以当A[i]比sum[i-1]大时，我们应该保存A[i]，忽略掉前面所有数的最大子数组和。因此，正确的DP方程应该是sum[i] = Math.max(A[i], sum[i-1] + A[i]).

AC Code

public class Solution {
    public int maxSubArray(int[] A) {
        if(A.length == 0) return 0;
        if(A.length == 1) return A[0];
        int [] sum = new int [A.length];
        //sum[i] store the max sum from A[0] to A[i]
        
        sum[0] = A[0];
        
        for(int i = 1; i < A.length; i++){
            sum[i] = Math.max(A[i], sum[i-1] + A[i]);
        }
        
        int maxSum = sum[0];
        for(int i = 0; i < A.length; i++){
            if(sum[i] > maxSum){
                maxSum = sum[i];
            }
        }
        
        return maxSum;
    }
}

但是上面的解法还是需要O（n）的额外存储空间，更加精简的code如下，直接维护maxLocal和maxGlobal，只需要O（1）的存储空间。

public class Solution {
    public int maxSubArray(int[] A) {
        if(A.length == 0) return 0;
        if(A.length == 1) return A[0];
        int maxLocal = A[0];
        int maxGlobal = A[0];
        
        for(int i = 1; i < A.length; i++){
            maxLocal = Math.max(A[i], maxLocal + A[i]);
            maxGlobal = Math.max(maxLocal, maxGlobal);
        }
        
        return maxGlobal;
    }
}