给你一个整数数组 nums
,请你找出一个具有最大和的连续子数组(子数组最少包含一个元素),返回其最大和。
子数组 是数组中的一个连续部分。
示例 1:
输入:nums = [-2,1,-3,4,-1,2,1,-5,4] 输出:6 解释:连续子数组 [4,-1,2,1] 的和最大,为 6 。
示例 2:
输入:nums = [1] 输出:1
示例 3:
输入:nums = [5,4,-1,7,8] 输出:23
提示:
1 <= nums.length <= 105
-104 <= nums[i] <= 104
进阶:如果你已经实现复杂度为 O(n)
的解法,尝试使用更为精妙的 分治法 求解。
设 dp[i]
表示 [0..i]
中,以 nums[i]
结尾的最大子数组和,状态转移方程 dp[i] = nums[i] + max(dp[i - 1], 0)
。
由于 dp[i]
只与子问题 dp[i-1]
有关,故可以用一个变量 f 来表示。
最大子数组和可能有三种情况:
- 在数组左半部分
- 在数组右半部分
- 跨越左右半部分
递归求得三者,返回最大值即可。
动态规划:
class Solution:
def maxSubArray(self, nums: List[int]) -> int:
res = f = nums[0]
for num in nums[1:]:
f = num + max(f, 0)
res = max(res, f)
return res
分治:
class Solution:
def maxSubArray(self, nums: List[int]) -> int:
def crossMaxSub(nums, left, mid, right):
lsum = rsum = 0
lmx = rmx = -inf
for i in range(mid, left - 1, -1):
lsum += nums[i]
lmx = max(lmx, lsum)
for i in range(mid + 1, right + 1):
rsum += nums[i]
rmx = max(rmx, rsum)
return lmx + rmx
def maxSub(nums, left, right):
if left == right:
return nums[left]
mid = (left + right) >> 1
lsum = maxSub(nums, left, mid)
rsum = maxSub(nums, mid + 1, right)
csum = crossMaxSub(nums, left, mid, right)
return max(lsum, rsum, csum)
left, right = 0, len(nums) - 1
return maxSub(nums, left, right)
动态规划:
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;
}
}
分治:
class Solution {
public int maxSubArray(int[] nums) {
return maxSub(nums, 0, nums.length - 1);
}
private int maxSub(int[] nums, int left, int right) {
if (left == right) {
return nums[left];
}
int mid = (left + right) >>> 1;
int lsum = maxSub(nums, left, mid);
int rsum = maxSub(nums, mid + 1, right);
return Math.max(Math.max(lsum, rsum), crossMaxSub(nums, left, mid, right));
}
private int crossMaxSub(int[] nums, int left, int mid, int right) {
int lsum = 0, rsum = 0;
int lmx = Integer.MIN_VALUE, rmx = Integer.MIN_VALUE;
for (int i = mid; i >= left; --i) {
lsum += nums[i];
lmx = Math.max(lmx, lsum);
}
for (int i = mid + 1; i <= right; ++i) {
rsum += nums[i];
rmx = Math.max(rmx, rsum);
}
return lmx + rmx;
}
}
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
}
public class Solution {
public int MaxSubArray(int[] nums) {
int res = nums[0], f = nums[0];
for (int i = 1; i < nums.Length; ++i)
{
f = nums[i] + Math.Max(f, 0);
res = Math.Max(res, f);
}
return res;
}
}
impl Solution {
pub fn max_sub_array(nums: Vec<i32>) -> i32 {
let n = nums.len();
let mut res = nums[0];
let mut sum = nums[0];
for i in 1..n {
let num = nums[i];
sum = num.max(sum + num);
res = res.max(sum);
}
res
}
}