Maximum Subarray
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Maximum Subarray
原题链接:https://leetcode.com/problems/maximum-subarray/description/
题目很简单,给定一个数组,找出一个子数组使得它的和最大,也就是找出该数组的最大连续和。
Solution1: 采取分治法来解决这个问题,首先,将原数组划分为两个元素个数相等的子数组,然后递归求解。注意,递归求解只能找出完全位于数组左半部或者右半部的最大连续和,所以我们还要找出横跨左半部和右半部两个子数组的最大连续和,并和左半部以及右半部的最大连续和进行比较。
class Solution {public: int maxSubArray(vector<int>& nums) { if (nums.size() == 1) return nums[0]; int m = nums.size()/2; vector<int> SubArray_A(nums.begin(), nums.begin() + m); vector<int> SubArray_B(nums.begin() + m, nums.end()); int largest_sum = max(maxSubArray(SubArray_A), maxSubArray(SubArray_B)); int max_A = INT_MIN, max_B = INT_MIN, s = 0; for (int i = SubArray_A.size() - 1; i >=0; i--) { s += SubArray_A[i]; max_A = max(s, max_A); } s = 0; for (int i = 0; i < SubArray_B.size(); i++) { s += SubArray_B[i]; max_B = max(s, max_B); } largest_sum = max(largest_sum, max_A + max_B); return largest_sum; }};
算法复杂度:当数组长度为n时,设T(n)为其时间复杂度,则有T(n) = 2T(n/2) + O(n), 所以,采用分治法,时间复杂度为O(nlogn)。
Solution2: 第一种方法的效率有点低,后来又改进了一下。
设置数组sum[n]来存储该序列的部分和,即sum[i]等于数组前i项的和。假设i < j,则sum[j] - sum[i]表示的就是我们要找的连续和,对于一个给定的j,要使sum[j] - sum[i]最大,那么,sum[i]就必须是最小的。所以,通过遍历数组sum,找出每一个sum[j]对应的最小的sum[i]就可以求解。
// Runtime: 9 ms// Your runtime beats 42.89 % of cpp submissions.class Solution {public: int maxSubArray(vector<int>& nums) { int n = nums.size(); int sum[n + 1]; sum[0] = 0; for (int i = 0; i < n; i++) sum[i + 1] = sum[i] + nums[i]; // largest sum = largest sum[i] - smallest sum[j], i > j int largest_sum = INT_MIN, smallest_sum_before_now = sum[0]; for (int i = 1; i <= n; i++) { largest_sum = (sum[i] - smallest_sum_before_now) > largest_sum ? (sum[i] - smallest_sum_before_now) : largest_sum; smallest_sum_before_now = (smallest_sum_before_now > sum[i]) ? sum[i] : smallest_sum_before_now; } return largest_sum; }};
算法复杂度:只有一层循环,复杂度为O(n)。
Soluntion3: Idea is very simple. Basically, keep adding each integer to the sequence until the sum drops below 0.
If sum is negative, then should reset the sequence.
class Solution {public: int maxSubArray(int A[], int n) { int ans=A[0],i,j,sum=0; for(i=0;i<n;i++){ sum+=A[i]; ans=max(sum,ans); sum=max(sum,0); } return ans; }};
算法复杂度:O(n)
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