[LeetCode] 53. Maximum Subarray 最大子数组 Given an integer array nums, find the contiguous
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.
这道题让求最大子数组之和，并且要用两种方法来解，分别是 O(n) 的解法，还有用分治法 Divide and Conquer Approach，这个解法的时间复杂度是 O(nlgn)，那就先来看 O(n) 的解法，定义两个变量 res 和 curSum，其中 res 保存最终要返回的结果，即最大的子数组之和，curSum 初始值为0，每遍历一个数字 num，比较 curSum + num 和 num 中的较大值存入 curSum，然后再把 res 和 curSum 中的较大值存入 res，以此类推直到遍历完整个数组，可得到最大子数组的值存在 res 中，代码如下：
c++ 解法一:
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;
}
};Java 解法一:
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;
}
}题目还要求我们用分治法 Divide and Conquer Approach 来解，这个分治法的思想就类似于二分搜索法，需要把数组一分为二，分别找出左边和右边的最大子数组之和，然后还要从中间开始向左右分别扫描，求出的最大值分别和左右两边得出的最大值相比较取最大的那一个，代码如下：
C++ 解法二:
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));
}
};Java 解法二:
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));
}
}到此这篇关于C++实现LeetCode(53.最大子数组)的文章就介绍到这了,更多相关C++实现最大子数组内容请搜索编程网以前的文章或继续浏览下面的相关文章希望大家以后多多支持编程网！
--结束END--
本文标题: C++实现LeetCode(53.最大子数组)
本文链接: https://www.lsjlt.com/news/130592.html(转载时请注明来源链接)
有问题或投稿请发送至: 邮箱/279061341@qq.com QQ/279061341
下载Word文档到电脑，方便收藏和打印～
2024-03-01
2024-03-01
2024-02-29
2024-02-29
2024-02-29
2024-02-29
2024-02-29
2024-02-29
2024-02-29
2024-02-29
回答
回答
回答
回答
回答
回答
回答
回答
回答
回答
官方手机版
微信公众号
商务合作
0