053 - 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.

int domerge(int *nums, int *nsize)
{
int kk = 0, k = *nsize, i;
if (nums[0] < 0) kk++;
for (i = kk+1; i <= k; i++) {
if (nums[i] < 0) {
if (i == k) break;
if (nums[i] + nums[kk] >= 0 && nums[i] + nums[i + 1] >= 0) {
nums[kk] += nums[i] + nums[i + 1];
i++;
} else {
nums[++kk] = nums[i];
}
} else {
if (nums[kk] >= 0)
nums[kk] += nums[i];
else
nums[++kk] = nums[i];
}
}

if (kk == *nsize) return 0;
*nsize = kk;
return 1;
}

int listmax(int *nums, int numsSize)
{
if (numsSize < 1) return 0;
if (numsSize == 1) return nums[0];
if (nums[0] <= 0) return nums[0] + listmax(nums + 1, numsSize - 1);
int nextgt0 = listmax(nums + 2, numsSize - 2);
if (nums[1] + nextgt0 >= 0)
return nums[0] + nums[1] + nextgt0;
else
return nums[0];
}
int maxSubArray(int *nums, int numsSize)
{
int max = INT_MIN;
int i, k = 0;
if (!numsSize) return 0;
/* merge +/- */
for( i = 0 ; i < numsSize ; i++ ) {
if (nums[i] > max) max = nums[i];
if (!i || !nums[i]) continue;
if (nums[i] >= 0) {
if (nums[k] >= 0) nums[k] += nums[i];
else nums[++k] = nums[i];
} else {
if (nums[k] > 0) nums[++k] = nums[i];
else nums[k] += nums[i];
}
}
if (k == 0) return nums[k] <= 0? max:nums[k];

/* merge +a -b +c --> a+b>=0 && b+c>=0 */
while (domerge(nums, &k))
;
max = INT_MIN;
int left = 0, right = k, sum;
while (left <= right) {
if (nums[left] <= 0) {
left++;
continue;
}
sum = listmax(nums + left, k+1-left);
if (sum > max) max = sum;
left++;
}
return max;
}