[Array]Minimum Size Subarray Sum
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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.
方法1:利用TwoPointer的思想进行遍历,时间复杂度为O(n)。
class Solution {public: int minSubArrayLen(int s, vector<int>& nums) { int start = 0,sum = 0,minlength=INT_MAX; for(int i=0;i<nums.size();++i){ sum+=nums[i]; while(sum>=s){ minlength = min(minlength,i-start+1); sum-=nums[start++]; } } return minlength==INT_MAX?0:minlength; }};
方法2:因为数组中的元素都是正数,那么对于数组中每一个位置上的元素,记录该位置前所有元素的累加和,并将这些累加和保存到一个vector中去,利用这个累加和vector进行操作,可以发现这个vector中的元素是递增性质,可以利用这个性质进行二分查找,这里我们可以发现accumulate[j]-accumulate[i]为i+1到j之间的累加和。正好满足subarray这个概念。这种方法的时间复杂度为O(nlog(n))。
class Solution {public: int minSubArrayLen(int s, vector<int>& nums) { vector<int> accumulate(nums.size()+1,0); int minLen = INT_MAX; int sum = 0; for(int i=1;i<accumulate.size();++i) accumulate[i]=accumulate[i-1]+nums[i-1]; for(int i=0;i<accumulate.size()-1;++i){ int target = s + accumulate[i]; int pos = upperbound(accumulate,target,i+1,accumulate.size()-1); if(pos==-1) continue; minLen = min(minLen,pos-i); } return minLen==INT_MAX?0:minLen; }private: int upperbound(vector<int>& accumulate,int target,int left,int right){ if(accumulate[right]<target) return -1; while(left<right){ int middle = (left+right)/2; if(accumulate[middle]<target) left=middle+1; else right=middle; } return left; }};
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