最大子数组和

maximum subarray (最大子数组和)

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Original ： Maximum Subarray

Question :

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

A subarray is a contiguous part of an array.

Sample :

Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.

Solution :

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we can solve the question using dp.
presume that the coordinate of the nums array range from \(0\) to \(n - 1\). \(dp[i]\) denoted that the maximum from \(0\) to \(i\).therefore , the formula is as follows:

\[f(i)=max(f(i−1)+nums[i],nums[i]) \]

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c++ code

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int MAX = -10000 - 1;
        int n = nums.size();
        vector<int> dp(n + 1,0);
        for(int i = 1;i <= n;i++){
            dp[i] = max(dp[i - 1] + nums[i - 1] , nums[i - 1]);
            MAX = MAX>dp[i]?MAX:dp[i];
        }
        return MAX;
    }
};

we noticed that this formula are only relative with \(i\) and \(i - 1\).so we can further reduce the memory used. namely that we use two variables for recording \(dp[i]\) and \(dp[i - 1]\) respectively. therefore c++ code modified are as follows:

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int MAX = -10000 - 1;
        int n = nums.size();
        int pre = 0,cur;
        for(int i = 0;i < n;i++){
            cur = max(pre + nums[i] , nums[i]);
            MAX = MAX>cur?MAX:cur;
            pre = cur;
        }
        return MAX;
    }
};