Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Example
Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.
Follow up
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
O(nlogn) runtime, O(logn) stack space – Divide and Conquer
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
|
public int (int[] nums) {
return subArrayHelper(nums, 0, nums.length - 1);
}
private int subArrayHelper(int[] nums, int left, int right){
if (left > right)
return Integer.MIN_VALUE;
int middle = (left + right) >> 1;
int leftSum = subArrayHelper(nums, left, middle - 1);
int rightSum = subArrayHelper(nums, middle + 1, right);
int sum = 0;
int sumFromLeft = 0;
for (int i = middle - 1; i >= left; i--){
sum += nums[i];
sumFromLeft = Math.max(sumFromLeft, sum);
}
sum = 0;
int sumFromRight = 0;
for (int i = middle + 1; i <= right; i++){
sum += nums[i];
sumFromRight = Math.max(sumFromRight, sum);
}
int middleSum = sumFromLeft + sumFromRight + nums[middle];
return Math.max(middleSum, Math.max(leftSum, rightSum));
}
|
O(n) runtime, O(1) space – Dynamic programming
1
2
3
4
5
6
7
8
9
10
11
|
public int (int[] nums) {
int[] dp = new int[nums.length+1];
int max = Integer.MIN_VALUE;
for (int i = 1; i <= nums.length; i++) {
dp[i] = Math.max(nums[i-1], dp[i-1] + nums[i-1]);
max = Math.max(max, dp[i]);
}
return max;
}
|
近期评论