【Leet Code】53. Maximum Subarray---Medium

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.

思路一：

遍历数组元素，每次加一个新的元素，如果当前和sum>之前保存的结果result，令result=sum；如果sum<0，令sum=0.最后result保存的就是最大的和值。

思路二：

就是实现题目中说到的divide and conquer approach。

1）将数组从mid分成left、right两部分，迭代找到leftMax和rightMax。令subMax=max（leftMax，rightMax）

2）然后是找mid连接的左右两边的元素的最大值。

先找mid左边的和最大值leftContigousMax（一定要包含nums[mid]，不然不是连续的和了），如果leftContigousMax就可以直接返回subMax了（因为这时可以肯定右边的最大值rightMax是最大的）。

同理，找到mid右边的和最大值rightContigousMax，如果rightContigousMax<0,直接返回subMax；

否则返回max（subMax，leftContigousMax+rightContigousMax）。

代码实现一：

class Solution {public:    int maxSubArray(vector<int>& nums) {        if(nums.size() < 1) return 0;        int result = INT_MIN;        int sum = 0;                for(auto num: nums)        {            sum += num;            if(sum > result) result = sum;            if(sum < 0) sum = 0;        }                return result;            }};

代码2实现：

class Solution {public:    int maxSubArray(vector<int>& nums) {        return maxSubArrayRec(nums, 0, nums.size() -1);    }private:int maxSubArrayRec(vector<int>&nums, int i, int j) {    if(i==j)        return nums[i];    int mid = (i + j) / 2;    int leftMax = maxSubArrayRec(nums,i,mid);    int rightMax = maxSubArrayRec(nums,mid+1,j);    int subMax = max(leftMax,rightMax);    int leftContiguousMax = nums[mid];    int leftSum = nums[mid];    for(int k=mid-1; k>=i; k--) {        leftSum += nums[k];        if(leftSum > leftContiguousMax)            leftContiguousMax = leftSum;    }    if(leftContiguousMax <= 0)        return subMax;    int rightContigousMax = nums[mid+1];    int rightSum = nums[mid+1];    for(int k=mid+2; k<=j; k++) {        rightSum += nums[k];        if(rightSum > rightContigousMax)            rightContigousMax = rightSum;    }    if(rightContigousMax <= 0)        return subMax;    return max(subMax,leftContiguousMax+rightContigousMax);}};