209. Minimum Size Subarray Sum解题报告
2017-10-10 21:21
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题目
Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous 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.
题目解析
题目大意是说给定一个数组,一个正整数s,找出最小连续的数组子集,使这个子集元素的和sum >= s, 如果这个数组所有元素的和比s小,则返回0思路分析
因为是为了找最短的子数组,因此可以一开始设定一个长度n,使得这个长度n的数组求和可以超过正整数s,于是以后遍历的时候,得到的数组长度最长只能是n。如果发现新的数组长度比n小,那么就更新数组长度n’,使以后遍历的数组最长不能超过n’,那么最后得到的n’就是最短长度#include <iostream> #include <vector> using namespace std; class Solution { public: int minSubArrayLen(int s, vector<int>& nums) { int _max = nums.size(); if (_max == 0) return 0; for (int i = 0; i < nums.size(); ++i) { if (i == 0) { int startmax = 0, result = 0, endpos = i; while (result < s && endpos < nums.size()) { result += nums[endpos]; startmax++; endpos++; } if (startmax < _max) _max = startmax; if (startmax == _max && result < s) return 0; } else { int addnum = _max, curaddnum = 1, result = nums[i]; while (curaddnum <= addnum){ if (result >= s) { if (curaddnum <= _max) { _max = curaddnum; break; } } if (i + curaddnum >= nums.size()) break; result += nums[i + curaddnum++]; } } } return _max; } }; int main() { int iarray[]={1,2,3,4,5}; size_t count=sizeof(iarray)/sizeof(int); vector<int> ivec3(iarray,iarray+count); Solution s; std::cout << s.minSubArrayLen(11, ivec3)<< std::endl; return 0; }
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