【leetcode】209. Minimum Size Subarray Sum
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题目:
Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead.
For example, given the array [2,3,1,2,4,3]
and s = 7
,
the subarray [4,3]
has the minimal length under the problem constraint.
翻译:
找出最小个数的连续子数组,使得它们的和大于等于s。
思路:
我的第一思路是动态规划,代码如下,可是大数据时MLE了。。。
//Memory Limit Exceededclass Solution {public:int minSubArrayLen(int s, vector<int>& nums) {vector<vector<int>> v;v.resize(nums.size());int length=0;for (int i=0;i<nums.size();i++){v[i].resize(nums.size());}for (int i=0;i<nums.size();i++){for (int j=i;j<nums.size();j++){if(i==j)v[i][j]=nums[i];elsev[i][j]=v[i][j-1]+nums[j];if (v[i][j]>=s){if(length==0)length=j-i+1;elselength=min(length,j-i+1);}}}return length;}};看了提示two points之后想到了另外一种思路,只要设定两个flag,i和j,两者的初始值都是0,j>=i。j一直向后移动,直到i,j之间的数字的和sum大等于s,之后i再开始自加,直到i,j之间的数字之和sum小于s,重复前面两步的过程,动态记录其间sum>=s时,最小的i,j距离。顺利AC。
class Solution {public:int minSubArrayLen(int s, vector<int>& nums) {int i=0,j=0;int result=0;int sum=0;for (;j<nums.size();j++){sum+=nums[j];if (sum>=s){for (;i>=0 && i<=j && sum>=s;){i++;sum-=nums[i-1];}if (result==0){result=j-i+2;}elseresult=min(result,j-i+2);}}return result;}};
结果
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