LeetCode:Maximum Subarray

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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.

直接想法遍历数组，累加过程中记录累加的出来的最大值，只要前若干项和小于0，则清空其值重新累加。

public class Solution {    public int maxSubArray(int[] A) {        int maxn=Integer.MIN_VALUE,res=0,len=A.length;                for(int i=0;i<len;i++)        {            res+=A[i];            maxn=Math.max(maxn,res);            if(res<0)res=0;        }     return maxn;    }}

其实还可以利用动态规划的思想，记录前n-1项累加和和当前项之和，并与当前项比较大小，记录大值(实质还是前若干项和小于0时抛弃重新计算)：

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

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