53.Maximum Subarray

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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.

click to show more practice.

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.

题意

找出一个和最大的子数组.注意这里的子数组是连续的.
可以用两种方法:
1. Kadane算法:O(n)
2. 分治法:O(nlogn)

Kadane算法

描述

Kadane算法属于动态规划
定义两个变量result和curSum，其中result保存最终要返回的结果，即最大的子数组之和，curSum初始值为0,是一个临时变量
每遍历一个数字num:
1. 比较curSum+num和num中的较大值存入curSum
2. 然后再把result和curSum中的较大值存入result

具体实现

class Solution {public:    int maxSubArray(vector<int>& nums) {        int result = INT_MIN, curSum = 0;        for (int i = 0; i < nums.size(); i++) {            curSum = max(curSum+nums[i], nums[i]);            result = max(result, curSum);        }        return result;    }};

复杂度分析

时间复杂度:O(n)
空间复杂度:O(1)

分治法

描述

分治法：最大子串和的区间有以下三种情况（low，high分别为左右边界，mid为(low+high)/2）：
(1) 区间完全在 A[low,mid-1]
(2) 区间完全在 A[mid+1,high]
(3) 区间包含有 A[mid], 从中间开始,往左右两侧扫描

具体实现

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

复杂度分析

时间复杂度:O(nlogn)
理解题目之后还是觉得此题没太大必要用分治法.

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