LeetCode--maximum-subarray

题目描述

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.

分析：迭代一遍数组，累加数组中的数字，当累加值sum<0时，重新开始累加。在这个过程中需要记录每次sum<0之前的最大累加值maxsum，最后输出maxsum。时间复杂度为O(N)

class Solution {
public:
    int maxSubArray(int A[], int n) {
    int sum = A[0], maxSum = A[0];
    for (int i = 1; i < n;i++)
    {
        if (sum < 0)
            sum = 0;//先判断之前的sum能否被这次利用（小于0则抛弃）
        sum += A[i];
        maxSum = max(maxSum, sum);
    }
    return maxSum;
    }
};

题目要求除了这个方法之外，还要求用分治法求解。

这个分治法的思想就类似于二分搜索法，我们需要把数组一分为二，分别找出左边和右边的最大子数组之和，然后还要从中间开始向左右分别扫描，求出的最大值分别和左右两边得出的最大值相比较取最大的那一个，代码如下：

class Solution {
public:
int maxSubArray(vector<int>& nums) {
if (nums.empty()) return 0;
return helper(nums, 0, (int)nums.size() - 1);
}
int helper(vector<int>& nums, int left, int right) {
if (left >= right) return nums[left];
int mid = left + (right - left) / 2;
int lmax = helper(nums, left, mid - 1);
int rmax = helper(nums, mid + 1, right);
int mmax = nums[mid], t = mmax;
for (int i = mid - 1; i >= left; --i) {
t += nums[i];
mmax = max(mmax, t);
}
t = mmax;
for (int i = mid + 1; i <= right; ++i) {
t += nums[i];
mmax = max(mmax, t);
}
return max(mmax, max(lmax, rmax));
}
};