Maximum Subarray

最新推荐文章于 2024-01-21 00:18:48 发布

小菜鸟的leetcode

最新推荐文章于 2024-01-21 00:18:48 发布

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文章标签： divide and conquer need another try

本文链接：https://blog.youkuaiyun.com/Scarlett_leetcode/article/details/51303458

本文介绍了一种使用分治法寻找给定数组中具有最大和的连续子数组的方法。通过递归地将问题分解为左右两部分，并计算跨越中间点的最大子数组和，最终找到全局最优解。

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Problem

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.

DP 来做 O（n）的复杂度是最好的，但是这个divide and conquer 解法也值得学习

class Solution {

    
    int helper(const vector<int>& nums, int left, int right ) {
        if(left > right) return 0;
        int mid = (left + right)/2;
        
        int leftRst = helper(nums, left, mid - 1);
        int rightRst = helper(nums, mid + 1, right);
        
        int leftSum = 0, curSum = 0;
        for( int i = mid - 1; i >= 0; i-- ) {
            curSum += nums[i];
            leftSum = max( curSum, leftSum);
        }
        
        int rightSum = 0; curSum = 0;
        for( int i = mid + 1; i <= right ; i++ ) {
            curSum += nums[i];
            rightSum = max( curSum, rightSum);
        }
        int crossRst = leftSum + rightSum + nums[mid];
        
        int rst = max(leftRst, rightRst);
        rst = max(rst, crossRst);
        
        return rst;
    }
public:
    int maxSubArray(vector<int>& nums) {
        return helper(nums, 0, nums.size() - 1);
    }
};