LeetCode 题解（172): Maximum Subarray

题目：

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.

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More practice:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

题解：

简单做法：全局sumMax记录全局最大，局部sum，当sum小于零时令sum = 0。

复杂做法：Divide and Conquer。左最大，右最大，左右+mid 连起最大。

C++版：

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        return maxSub(nums, 0, nums.size() - 1);
    }
    
    int maxSub(vector<int>& nums, int start, int end) {
        if(start > end)
            return INT_MIN;
        int mid = (start + end) / 2;
        int leftMax = maxSub(nums, start, mid - 1);
        int rightMax = maxSub(nums, mid + 1, end);
        int sum = 0, lmax = 0, rmax = 0;
        for(int i = mid - 1; i >= start; i--) {
            sum += nums[i];
            if(sum > lmax)
                lmax = sum;
        }
        sum = 0;
        for(int i = mid + 1; i <= end; i++) {
            sum += nums[i];
            if(sum > rmax)
                rmax = sum;
        }
        return max(max(leftMax, rightMax), lmax + rmax + nums[mid]);
    }
};

Java版：

public class Solution {
    public int maxSubArray(int[] nums) {
        return maxSub(nums, 0, nums.length - 1);
    }
    
    public int maxSub(int[] nums, int start, int end) {
        if(start > end)
            return Integer.MIN_VALUE;
        int mid = (start + end) / 2;
        int leftMax = maxSub(nums, start, mid - 1);
        int rightMax = maxSub(nums, mid + 1, end);
        int sum = 0, lmax = 0, rmax = 0;
        for(int i = mid - 1; i >= start; i--) {
            sum += nums[i];
            if(sum > lmax)
                lmax = sum;
        }
        sum = 0;
        for(int i = mid + 1; i <= end; i++) {
            sum += nums[i];
            if(sum > rmax)
                rmax = sum;
        }
        return Math.max(Math.max(leftMax, rightMax), lmax + nums[mid] + rmax);
    }
}

Python版：

class Solution:
    # @param {integer[]} nums
    # @return {integer}
    def maxSubArray(self, nums):
        maxSum, sum, i = -sys.maxint, 0, 0
        while i < len(nums):
            sum += nums[i]
            if sum > maxSum:
                maxSum = sum
            if sum < 0:
                sum = 0
            i += 1
        return maxSum