【LeetCode & 劍指offer刷題】動態規劃與貪婪法題4:42 連續子數組的最大和(53. Maximum Subarray)
阿新 • • 發佈:2019-01-06
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Follow up:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
//最大子數組問題
/*
方法:動態規劃
f(i) = a[i], i <0或f(i-1) <= 0;
f(i) = f(i-1) + a[i], i!=0, f(i-1) > 0
max(f[i]) 不太好想
*/
/*方法一:動態規劃
sum代表了包含nums[i]時(以nums[i]結尾的子數組)的最大和
不斷更新sum和res sum = max(sum + num, num)
分析:時間復雜度O(n)(線性)
*/
class Solution
{
public:
int maxSubArray(vector<int>& nums)
{
int res = INT_MIN, sum = 0;
for (int num : nums)
{
sum = max(sum + num, num);
res = max(res, sum);
}
return res;
}
};
//方法二:分治法
//具體過程:將數組分為左子數組、右子數組、跨越中點的子數組問題
//分析:時間復雜度O(nlgn)
//參考資料:《算法導論》
class Solution
{
public:
int maxSubArray(vector<int>& a)
{
return findMaxSubArray(a, 0, a.size()-1);//遞歸入口
}
//遞歸函數:找a[left...right]的最大子數組(歸並排序和快速排序中也用到了分治法,可以類比一下)
int findMaxSubArray(vector<int>& a, int left, int right)//遞歸函數
{
if(right == left) return a[left];
int mid = (left+right)/2;
int left_sum = findMaxSubArray(a, left, mid);
int right_sum = findMaxSubArray(a, mid+1, right);
int cross_sum = findMaxCrossingSubArray(a, left, mid, right);
return max(max(left_sum, right_sum), cross_sum);
}
//找跨中點的最大子數組,我們只需找出形如A[i.. mid] 和A[mid+ 1. .j] 的最大子數組,然後將其合並即可。
int findMaxCrossingSubArray(vector<int>& a, int left, int mid, int right)
{
int left_sum,sum,right_sum;
sum = 0;
left_sum = a[mid];//初始化為參與計算的第一個元素
for(int i = mid; i>=left; i--)//從中間往左邊遍歷
{
sum += a[i];
if(sum>left_sum) left_sum = sum;
}
sum = 0;
right_sum = a[mid+1];//初始化
for(int j = mid+1; j<=right; j++) //從中間往右邊遍歷
{
sum += a[j];
if(sum>right_sum) right_sum = sum;
}
return (left_sum+right_sum);
}
};
【LeetCode & 劍指offer 刷題筆記】目錄(持續更新中...)
53. Maximum Subarray
Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. Example: Input: [-2,1,-3,4,-1,2,1,-5,4], Output: 6 Explanation: [4,-1,2,1] has the largest sum = 6.
【LeetCode & 劍指offer刷題】動態規劃與貪婪法題4:42 連續子數組的最大和(53. Maximum Subarray)