[leetcode]: 53. Maximum Subarray

1.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.

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

求连续子数组的最大和。
尝试用分治法解决它。

2.分析

用cur记录当前子数组的和
用max记录历史最大值
以[-2,1,-3,4,-1,2,1,-5,4]为例，遍历一遍看

x=-2
cur=-2,max=-2
x=1
cur=1,max=1 (-2+1<1, cur=-2的贡献是反作用的,所以要舍弃-2)
x=-3
cur=-2,max=1 (cur=cur+x=1-3=-2)
x=4
cur=4,max=4 (-2+4<4, cur=-2的贡献是反作用的,所以要舍弃-2)
x=-1
cur=3,max=4 (cur=cur+x=4-1=3)
x=2
cur=5,max=5 (cur=cur+x=3+2=5)
x=1
cur=6,max=6 (cur=cur+x=5+1=6)
x=-5
cur=1,max=6 (cur=cur+x=6-5=1)
x=4
cur=5,max=6 (cur=cur+x=1+5)

因此，如果前面计算的连续子数组的和cur<0, 就舍弃。从当前位置重新开始。
动态规划状态转移方程
dp[i],i表示以位置i为结尾的连续子数组的和的最大值
dp[i]=dp[i-1]+nums[i] dp[i-1]>0;
dp[i]=nums[i], dp[i-1]<=0;

3.代码

class Solution {public:    int maxSubArray(vector<int>& nums) {        int max_cur = 0;        int max_sofar = INT_MIN;        for (int n : nums) {            max_cur = max_cur > 0 ? max_cur + n : n;            max_sofar = max_cur > max_sofar ? max_cur : max_sofar;        }        return max_sofar;    }    //动态规划    int maxSubArray(vector<int>& nums) {        int* dp = new int[nums.size()]();        dp[0] = nums[0];        for (int i = 1; i < nums.size(); i++) {            dp[i] = dp[i - 1] > 0 ? dp[i - 1] + nums[i] : nums[i];        }        int maxP = INT_MIN;        for (int i = 0; i < nums.size(); i++)            maxP = dp[i] > maxP ? dp[i] : maxP;        return maxP;    }};