Leetcode53 Maximum Subarray

最新推荐文章于 2021-12-20 16:36:02 发布

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本文探讨了寻找具有最大和的连续子数组的问题，提供了一种暴力遍历解决方案和一种更高效的动态规划方法。动态规划方法将时间复杂度从1340ms降低到了4ms，显著提高了效率。

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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:

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.

暴力遍历，1340ms

//3 temp
class Solution {
public:
	int maxSubArray(vector<int>& nums) {
		int len = nums.size();
		vector<int> temp(len,0);
		int maxval = nums[0];
		for (int i = 0; i < len; i++) {
			temp[i] = nums[i];
			if (temp[i]>maxval) {
				maxval = temp[i];
			}
		}
		for (int i = 0; i < len; i++) {
			vector<int> temp(len, nums[i]);
			for (int j = i+1; j < len; j++) {
				temp[j]= temp[j - 1] + nums[j];
				if (temp[j] > maxval) {
					maxval = temp[j];
				}
			}
		}
		return maxval;
	}
};

动态规划4ms

//3 temp
class Solution {
public:
	int maxSubArray(vector<int>& nums) {
		int len = nums.size(),maxval=nums[0];
		vector<int> dp(len, nums[0]);
		for (int i = 1; i < len;i++) {
			if (dp[i - 1]>0) {
				dp[i] = dp[i - 1] + nums[i];
				maxval = max(dp[i],maxval);
			}
			else
			{
				dp[i] = nums[i];
				maxval = max(dp[i], maxval);
			}
		}

		return maxval;
	}
};