forked from doocs/leetcode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy pathSolution.cpp
74 lines (70 loc) · 1.67 KB
/
Solution.cpp
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
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
/*
* @lc app=leetcode.cn id=53 lang=cpp
*
* [53] 最大子序和
*
* https://leetcode-cn.com/problems/maximum-subarray/description/
*
* algorithms
* Easy (46.59%)
* Likes: 2144
* Dislikes: 0
* Total Accepted: 270.6K
* Total Submissions: 524.4K
* Testcase Example: '[-2,1,-3,4,-1,2,1,-5,4]'
*
* 给定一个整数数组 nums ,找到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。
*
* 示例:
*
* 输入: [-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。
*
*
* 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。
*
*/
#include <vector>
using std::vector;
// 贪心
class Solution0 {
public:
int maxSubArray(vector<int>& nums) {
if (nums.size() == 0) {
return 0;
}
int result = nums[0];
int temp = nums[0];
for (int i = 1; i < nums.size(); i++) {
if (temp >= 0) {
temp += nums[i];
} else {
temp = nums[i];
}
result = std::max(result, temp);
}
return result;
}
};
// @lc code=start
// 动态规划
class Solution {
public:
int maxSubArray(vector<int>& nums) {
if (nums.size() == 0) {
return 0;
}
vector<int> dp = vector<int>(nums.size());
dp[0] = nums[0];
int result = nums[0];
for (int i = 1; i < nums.size(); i++) {
dp[i] = std::max(dp[i - 1] + nums[i], nums[i]);
result = std::max(dp[i], result);
}
return result;
}
};
// @lc code=end