LeetCode() Minimun Size Subarray Sum

别人的代码

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
         int l, r, cum, res = nums.size()+1;
    l = r = cum = 0;
    while ((unsigned int)r < nums.size()) {
        cum += nums[r++];
        while (cum >= s) {
            res = min(res, r-l);
            cum -= nums[l++];
        }
    }
    return res<=nums.size()?res:0;
    }
};

我的 280ms

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        vector<int> res;
        int start=0,end=0,len=INT_MAX;
        for(int end=0;end<nums.size();++end)
        {
             while(sum(nums,s,start,end) && start<=end)
             {
                 (end-start+1 < len)? len=end-start+1:len;
                 start++;
             }
        }
        if(len == INT_MAX)
            return 0;
        return len;
    }
    bool sum(vector<int>& nums,int s,int i,int j)
    {
        for(int k=i;k<=j;++k)
            s=s-nums[k];
        return s<=0;
    }
};

nlogn的解法：把数组依次相加，然后用二分查找法找到sum-nums[i].不高效，但是一个思路。

int minSubArrayLen(int s, vector<int>& nums) {
        vector<int> sums = accumulate(nums);
        int n = nums.size(), minlen = INT_MAX;
        for (int i = 1; i <= n; i++) {
            if (sums[i] >= s) {
                int p = upper_bound(sums, 0, i, sums[i] - s);
                if (p != -1) minlen = min(minlen, i - p + 1);
            }
        }
        return minlen == INT_MAX ? 0 : minlen;
    }
private:
    vector<int> accumulate(vector<int>& nums) {
        int n = nums.size();
        vector<int> sums(n + 1, 0);
        for (int i = 1; i <= n; i++)
            sums[i] = nums[i - 1] + sums[i - 1];
        return sums;
    }
    int upper_bound(vector<int>& sums, int left, int right, int target) {
        int l = left, r = right;
        while (l < r) {
            int m = l + ((r - l) >> 1);
            if (sums[m] <= target) l = m + 1;
            else r = m;
        }
        return sums[r] > target ? r : -1;
    }