209. Minimum Size Subarray Sum

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.

给定一组数和一个值，找到最短的子串，使得子串的和大于等于给定值。

我的思路：

模拟双指针的操作：

class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int n=nums.size();
if(n==0) return 0;
int i=0;
int j=0;
int sum=0;
int d=numeric_limits<int>::max();
while(j<n)
{
while(sum<s&&j<n)
sum+=nums[j++];
if(j==n&&sum<s) return d==numeric_limits<int>::max()?0:d;
j--;
d=min(d,j-i+1);//j指向当前最后一个元素，i指向第一个元素
if(d==1) break;
sum-=nums[i++];
if(j==n-1)//j已经到末尾了
{
while(sum-nums[i]>=s)
{
sum-=nums[i++];
d=min(d,j-i+1);
}
break;
}
if(j<n-1)
{
j++;
sum-=nums[i++];
while(sum-nums[i]+nums[j]>=s)
{
sum-=nums[i++];
d=min(d,j-i+1);
}
}

}
return d<=n?d:0;
}
};

使用双指针较为简洁的方法：

class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int n=nums.size();
if(n==0) return 0;
int left=0;
int sum=0;
int d=numeric_limits<int>::max();
for(int i=0;i<n;i++)
{
sum+=nums[i];
while(sum>=s)
{
d=min(d,i+1-left);
sum-=nums[left++];
}
}
return d==numeric_limits<int>::max()?0:d;
}
};

另外一种二分法，虽然复杂度是nlogn的但是，也值得掌握

明天看吧：

int minSubArrayLen(int s, vector<int>& nums)
{
int n = nums.size();
if (n == 0)
return 0;
int ans = INT_MAX;
vector<int> sums(n + 1, 0); //size = n+1 for easier calculations
//sums[0]=0 : Meaning that it is the sum of first 0 elements
//sums[1]=A[0] : Sum of first 1 elements
//ans so on...
for (int i = 1; i <= n; i++)
sums[i] = sums[i - 1] + nums[i - 1];
for (int i = 1; i <= n; i++) {
int to_find = s + sums[i - 1];
auto bound = lower_bound(sums.begin(), sums.end(), to_find);
if (bound != sums.end()) {
ans = min(ans, static_cast<int>(bound - (sums.begin() + i - 1)));
}
}
return (ans != INT_MAX) ? ans : 0;
}