[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.

click to show more practice.

More practice:

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

Solution: dp

public class Solution {
    public int maxSubArray(int[] A) {
        int sum=0;
        int max=Integer.MIN_VALUE;
        for(int i=0;i<A.length;++i){
            sum+=A[i];
            if(sum>=max)
                max=sum;
            if(sum<0)
                sum=0;
        }
        return max;   
    }
}

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20150219:

通过维护一个全局最优和一个局部最优来做此题。

全局最优：到当前元素为止的最优解；

局部最优：必须包含当前元素的最优解。

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对每一个A[i]，都要判断A[i]是自立门户单干，还是跟着前边儿的元素一起。

也就是

localMax=Math.max(localMax+A[i],A[i]);

这句代码的实现的。

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