LeetCode Maximum Subarray

LeetCode Maximum Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.

click to show more practice.

More practice:If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.


求最大连续子数组和,这是一道非常经典的题目,从本科时候就开始做了。有两种解法。

动态规划法

用dp数组存储到当前元素的最大子数组和,如dp[i-1]表示包含第i-1个元素的最大子数组和,则dp[i]表示包含第i个元素的最大子数组和。如果dp[i-1]>0,则dp[i]=dp[i-1]+nums[i];否则dp[i]=nums[i]。完整代码如下:

class Solution {
public:
	int maxSubArray(vector<int>& nums) {
		vector<int> dp(nums.size(), 0);
		dp[0] = nums[0];
		int ans = nums[0];
		for (int i = 1; i < nums.size(); i++) {
			if (dp[i - 1] > 0)dp[i] = dp[i - 1] + nums[i];
			else dp[i] = nums[i];
			if (dp[i] > ans)ans = dp[i];
		}
		return ans;
	}
};

本代码只用扫描一遍数组,所以时间复杂度为O(n)。本代码提交AC,用时12MS。

分治法

将数组一分为二,比如数组[a,b]分为[a,m-1]和[m+1,b],递归求解[a,m-1]和[m+1,b]的最大连续子数组和lmax和rmax,但是[a,b]的最大连续子数组和还可能是跨过中点m的,所以还应该从m往前和往后求一个包含m的最大连续子数组和mmax,然后再求lmax,rmax,mmax的最大值。完整代码如下:

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

本代码可以类比平面上求最近点对的例子,需要考虑跨界的问题,时间复杂度为O(nlgn)。本代码提交AC,用时也是12MS。

综合来看,还是动态规划法更漂亮。

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  1. Pingback: LeetCode Maximum Product Subarray | bitJoy > code

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