leetcode 53. 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.

也是很经典的老题啦。

经典问题经典解法：

public int maxSubArray(int[] nums) {int maxSum=Integer.MIN_VALUE;int sum=0;for(int i=0;i<nums.length;i++){if(sum<=0){sum=nums[i];}else{sum=sum+nums[i];}if(sum>maxSum){maxSum=sum;}}return maxSum;}

这个经典的解法大概思路是这样的：

sum(0,i) = a[i] + (sum(0,i-1) < 0 ? 0 : sum(0,i-1))

下面是大神的代码，跟我的代码很类似，思路也相同。

int maxSubArray(vector<int>& nums) {        int n=nums.size();        int max=nums[0],sum=nums[0];        for(int i=1;i<n;i++){            sum=((sum<0)?0:sum)+nums[i];            max=sum>max?sum:max;        }     return max;}

还有大神想到用DP来做。

把子问题看做为: maxSubArray(int A[], int i), 代表 以第 i 个元素结尾 的子数组 A[ 0:i ]最大和。关系公式如下：

maxSubArray(A, i) = ( maxSubArray(A, i - 1) > 0 ? maxSubArray(A, i - 1) : 0 ) + A[i]; 

其实也是跟经典解法差不多的思想嘛。

public int maxSubArray(int[] A) {        int n = A.length;        int[] dp = new int[n];//dp[i] means the maximum subarray ending with A[i];        dp[0] = A[0];        int max = dp[0];                for(int i = 1; i < n; i++){            dp[i] = A[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0);            max = Math.max(max, dp[i]);        }                return max;}