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public class Solution {

/**

* @param nums: A list of integers

* @return: A integer indicate the sum of max subarray

*/

public int maxSubArray(ArrayList<Integer> nums) {

// write your code

// brute force O(n^2)

if (nums == null) {

return 0;

}

int max = nums.get(0);

int size = nums.size();

for (int i = 0; i < size; i++) {

int tempSum = nums.get(i);

max = max < tempSum? tempSum : max;

for (int j = i + 1; j < size; j++) {

if (tempSum > 0) {

tempSum += nums.get(j);

max = max < tempSum? tempSum : max;

} else {

break;

}

}

}

return max;

}

}

// O(n) solution.

public class Solution {

/**

* @param nums: A list of integers

* @return: A integer indicate the sum of max subarray

*/

public int maxSubArray(ArrayList<Integer> nums) {

// write your code

if (nums == null) {

return 0;

}

int max = nums.get(0);

int previousSum = max;

int size = nums.size();

// as long as previous sum is positive, it can

// contribute to following subarray. however

// negative values in the folloing subarry

// might decrease the totalsum, so we need to

// update the global sum with possible

// maximal sum.

for (int i = 1; i < size; i++) {

int current = nums.get(i);

previousSum = Math.max(current, current + previousSum);

max = max < previousSum? previousSum : max;

}

return max;

}

}