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Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Example 1:
Input: nums = [-2,1,-3,4,-1,2,1,-5,4] Output: 6 Explanation: [4,-1,2,1] has the largest sum = 6.
Example 2:
Input: nums = [1] Output: 1
Example 3:
Input: nums = [5,4,-1,7,8] Output: 23
Constraints:
1 <= nums.length <= 3 * 104-105 <= nums[i] <= 105Follow up: If you have figured out the
O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
class Solution:
def maxSubArray(self, nums: List[int]) -> int:
n = len(nums)
res = f = nums[0]
for i in range(1, n):
f = nums[i] + max(f, 0)
res = max(res, f)
return res
class Solution {
public int maxSubArray(int[] nums) {
int f = nums[0], res = nums[0];
for (int i = 1, n = nums.length; i < n; ++i) {
f = nums[i] + Math.max(f, 0);
res = Math.max(res, f);
}
return res;
}
}
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int f = nums[0], res = nums[0];
for (int i = 1; i < nums.size(); ++i) {
f = nums[i] + max(f, 0);
res = max(res, f);
}
return res;
}
};
/**
* @param {number[]} nums
* @return {number}
*/
var maxSubArray = function (nums) {
let f = nums[0],
res = nums[0];
for (let i = 1; i < nums.length; ++i) {
f = nums[i] + Math.max(f, 0);
res = Math.max(res, f);
}
return res;
};
func maxSubArray(nums []int) int {
f, res := nums[0], nums[0]
for i := 1; i < len(nums); i++ {
if f > 0 {
f += nums[i]
} else {
f = nums[i]
}
if f > res {
res = f
}
}
return res
}
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