剑指 Offer 42. 连续子数组的最大和

思路#

 方法一：动态规划#

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        //memo[i]表示以nums[i]结尾并且包含nums[i]的子数组的最大和
        vector<int> memo(nums.size());
        memo[0] = nums[0];

        for(int i = 1; i < nums.size(); ++i) {
            if(memo[i-1] > 0)
                memo[i] = memo[i-1] + nums[i];
            else
                memo[i] = nums[i];
        }

        int maxSum = memo[0];
        for(int i = 1; i < memo.size(); ++i) {
            if(memo[i] > maxSum)
                maxSum = memo[i];
        }

        return maxSum;
    }
};

复杂度分析#

时间复杂度：O(n)

空间复杂度：O(n)

 

 方法二：模拟#

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int maxSum = nums[0];
        int thisSum = nums[0];

        for(int i = 1; i < nums.size(); ++i) {
            if(thisSum < 0)
                thisSum = 0;

            thisSum += nums[i];
            if(thisSum > maxSum)
                maxSum = thisSum;
        }

        return maxSum;
    }
};

复杂度分析#

时间复杂度：O(n)

空间复杂度：O(1)