【刷题-LeetCode】209. Minimum Size Subarray Sum

1. 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.

Example:

Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.

Follow up:

If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).

解法1 暴力搜索

找出所有的$\sum_{k=i}^j a_k \geq s \(的子串，取长度\)j-i+1$最小的

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        if(nums.size() == 0)return 0;
        int ans = INT_MAX;
        for(int i = 0; i < nums.size(); ++i){
            int tmp_sum = 0;
            for(int j = i; j < nums.size() && j - i <= ans; ++j){
                tmp_sum += nums[j];
                if(tmp_sum >= s)ans = min(ans, j-i+1);
            }
        }
        return ans == INT_MAX ? 0 : ans;
    }
};

Note :

• 在内层循环中，一定要加j - i <= ans的判断条件，否则会超时
• 为了避免在内层循环中重复求和，可以先计算nums的前n项和，放到数组sum中

解法2 二分查找。前n项和数组sum一定是单调递增的，原问题可以转换为：

查找\(sum[i] + s\)在sum中第一次出现的位置\((i = 0, 1, 2, ..., nums.size())\)，即找\(nums[i] + ... + nums[j] \geq s\)对应的最小长度

直接调用c++ stl中的lower_bound()函数

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        if(nums.size() == 0)return 0;
        vector<int>sum(nums.size() + 1, 0);
        for(int i = 0; i < nums.size(); ++i)sum[i+1] = sum[i]+nums[i];
        int ans = INT_MAX;
        for(int i = 1; i <= nums.size(); ++i){
            int to_find = s + sum[i-1];
            auto bound = lower_bound(sum.begin(), sum.end(), to_find);
            if(bound != sum.end())ans = min(ans, int(bound - sum.begin()) - i + 1);
        }
        return ans == INT_MAX ? 0 : ans;
    }
};

解法3 one-pass。记录满足\(sum \geq s\)的子串的起始位置，在找到一个符合条件的子串后，不断收缩子串

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