leetcode53 Maximum Subarray

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.

最普通的一道最大连续子段和。动态规划的经典问题。

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

注意maxSubSum要初始化成nums[0]而不是0。

优化的解法：

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

因为此题不需要求出最大子段和区间，所以无需dp数组，一次遍历就可以求出。