leetcode 53. 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.

 

分析：求连续子数组的最大和

思路一：动态规划，定义的数组dp[], dp[i]表示以nums[i]结尾的最大子数组的和，计算dp[i + 1]时, dp[i + 1] = max(nums[i + 1], nums[i + 1] + dp[i])，如果dp[i]小于0,那么dp[i + 1] = nums[i + 1].

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

思路二：分治法, 以包括或者不包括数组中间元素作为分治的条件。

class Solution {
    int recursion(vector<int> nums, int l, int r) {
        if (l > r)
            return INT_MIN;
        if (l == r)
            return nums[l];
        int mid = ((r - l) >> 1) + l;
        //int mid = (l + r) / 2;
        int left = recursion(nums, l, mid - 1);
        int right = recursion(nums, mid + 1, r);
        int t = nums[mid], maxn = nums[mid];
        for (int i = mid - 1; i >= l; i--) {
            t += nums[i];
            maxn = max(maxn, t);
        }
        t = maxn;
        for (int i = mid + 1; i <= r; i++) {
            t += nums[i];
            maxn = max(maxn, t);
        }
        return max(max(left, right), maxn);
    }
public:
    int maxSubArray(vector<int>& nums) {
        int len = nums.size();
        if (len == 0)
            return 0;
        return recursion(nums, 0, len - 1);
    }
};