zoukankan      html  css  js  c++  java
  • 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.

    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). 

    Approach #1:

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

    Runtime: 8 ms, faster than 98.81% of C++ online submissions for Minimum Size Subarray Sum.

    Approach #2: Using two pointer:

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

    Runtime: 8 ms, faster than 98.81% of C++ online submissions for Minimum Size Subarray Sum.

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    c# 复制整个文件夹的内容,Copy所有文件
    c# 创建文件夹
    c# 访问共享文件
    sublimit 编辑器 设置默认的编码
    WPF xml配置文件里面的大于小于号转义
    c# datatable 分组
    WPF 耗时操作时,加载loging 动画 (BackgroundWorker 使用方法)
    WPF DEV gridcontrol 自定义计算列(TotalSummary)
    postgresql 创建gin索引
    WPF DEV gridcontrol当前项的数据导出为mdb文件
  • 原文地址:https://www.cnblogs.com/h-hkai/p/9890551.html
Copyright © 2011-2022 走看看