/**
*
* @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。