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.

 

给定一个数组，求元素和最大的子数组。

解法1： 动态规划DP，dp[i] 表示到元素i时，末尾为i的数组的最大和。dp[i-1] 的最大元素和如果为正，那么有用就和当前元素相加。如果为负那就无用，干脆舍弃，只用dp[i] 自己来组成。

解法2：题中让用divide and conquer的方法来做。

Java:

public class Solution {
public int maxSubArray(int[] nums) {
int res = Integer.MIN_VALUE, curSum = 0;
for (int num : nums) {
curSum = Math.max(curSum + num, num);
res = Math.max(res, curSum);
}
return res;
}
}

Java: Divide and conquer

public class Solution {
public int maxSubArray(int[] nums) {
if (nums.length == 0) return 0;
return helper(nums, 0, nums.length - 1);
}
public int helper(int[] nums, int left, int right) {
if (left >= right) return nums[left];
int mid = left + (right - left) / 2;
int lmax = helper(nums, left, mid - 1);
int rmax = helper(nums, mid + 1, right);
int mmax = nums[mid], t = mmax;
for (int i = mid - 1; i >= left; --i) {
t += nums[i];
mmax = Math.max(mmax, t);
}
t = mmax;
for (int i = mid + 1; i <= right; ++i) {
t += nums[i];
mmax = Math.max(mmax, t);
}
return Math.max(mmax, Math.max(lmax, rmax));
}
}

Python:

class Solution(object):
def maxSubArray(self, nums):
if max(nums) < 0:
return max(nums)
global_max, local_max = 0, 0
for x in nums:
local_max = max(0, local_max + x)
global_max = max(global_max, local_max)
return global_max

Python: wo

class Solution(object):
def maxSubArray(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
dp = [0] * (len(nums) + 1)
max_sum = float('-inf')  # cannot set max_sum = 0, error when input is [-1]
for i in xrange(1, len(nums) + 1):
if dp[i-1] >= 0:
dp[i] = dp[i-1] + nums[i-1]
else:
dp[i] = nums[i-1]
max_sum = max(max_sum, dp[i])

return max_sum 　　

Python: wo

class Solution(object):
def maxSubArray(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
dp = [0] * len(nums)
# res = float('-inf') # error [1]
dp[0] = nums[0]
res = dp[0]
for i in xrange(1, len(nums)):
dp[i] = nums[i] + max(0, dp[i-1])
res = max(res, dp[i])

return res 　　

C++:

class Solution {
public:
int maxSubArray(vector<int>& nums) {
int res = INT_MIN, curSum = 0;
for (int num : nums) {
curSum = max(curSum + num, num);
res = max(res, curSum);
}
return res;
}
};

 C++: Divide and conquer

class Solution {
public:
int maxSubArray(vector<int>& nums) {
if (nums.empty()) return 0;
return helper(nums, 0, (int)nums.size() - 1);
}
int helper(vector<int>& nums, int left, int right) {
if (left >= right) return nums[left];
int mid = left + (right - left) / 2;
int lmax = helper(nums, left, mid - 1);
int rmax = helper(nums, mid + 1, right);
int mmax = nums[mid], t = mmax;
for (int i = mid - 1; i >= left; --i) {
t += nums[i];
mmax = max(mmax, t);
}
t = mmax;
for (int i = mid + 1; i <= right; ++i) {
t += nums[i];
mmax = max(mmax, t);
}
return max(mmax, max(lmax, rmax));
}
};

 

类似题目：

[LeetCode] 70. Climbing Stairs 爬楼梯

[Airbnb] Max Sum of Non-consecutive Array Elements

　　

All LeetCode Questions List 题目汇总