/**
*
* @author gentleKay
* 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.
*
*
* 例如，给定数组[−2,1、−3,4、−1,2,1、−5,4]，
* 相邻子阵列[4、−1、2、1]的最大和=6。
* 单击此处可显示更多练习。
* 更多实践：
* 如果您已经找到了O（N）解决方案，请尝试使用分而治之的方法对另一个解决方案进行编码，这更为微妙。
*/
/**
*
* @author gentleKay
* 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.
*
*
* 例如，给定数组[−2,1、−3,4、−1,2,1、−5,4]，
* 相邻子阵列[4、−1、2、1]的最大和=6。
* 单击此处可显示更多练习。
* 更多实践：
* 如果您已经找到了O（N）解决方案，请尝试使用分而治之的方法对另一个解决方案进行编码，这更为微妙。
*/
public class Main17 {
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] A = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
System.out.println(Main17.maxSubArray(A));
}
public static int maxSubArray(int[] A) {
int max = Integer.MIN_VALUE;
for (int i=0;i<A.length;i++) {
int sum = 0;
for (int j=i;j<A.length;j++) {
sum += A[j];
if(sum > max){
max = sum;
}
}
}
return max;
}
}
解题思路：
主要是进行循环遍历，相加的值与MAX相比较，选出最大的MAX即可。