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  • Minimum Size Subarray Sum

    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 -1 instead.

    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.

    Challenge 

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

    Analyse: Two pointers. O(n)

    Runtime: 203ms

     1 class Solution {
     2 public:
     3     /**
     4      * @param nums: a vector of integers
     5      * @param s: an integer
     6      * @return: an integer representing the minimum size of subarray
     7      */
     8     int minimumSize(vector<int> &nums, int s) {
     9         // write your code here
    10         // mark left, set right as left + 1
    11         // move right until sum of nums[left...right] >= s
    12         // update the min length and let left move one position
    13         // if current sum >= s update min length and keep move left
    14         // if current sum < s move right one position
    15         
    16         if (nums.empty()) return -1;
    17         int left = 0, right = 1;
    18         int minSize = INT_MAX, tempSum = nums[left];
    19         if (tempSum > s) return 1;
    20         while (left < nums.size() && right < nums.size()) {
    21             tempSum += nums[right];
    22             while (tempSum >= s) {
    23                 minSize = min(minSize, right - left + 1);
    24                 tempSum -= nums[left];
    25                 left++;
    26             }
    27             if (tempSum < s) {
    28                 right++;
    29             }
    30         }
    31         return minSize == INT_MAX ? -1 : minSize;
    32     }
    33 };

    Analyse: Brute force. O(n^2)

    Runtime: 605ms

     1 class Solution {
     2 public:
     3     /**
     4      * @param nums: a vector of integers
     5      * @param s: an integer
     6      * @return: an integer representing the minimum size of subarray
     7      */
     8     int minimumSize(vector<int> &nums, int s) {
     9         // write your code here
    10         if (nums.empty()) return -1;
    11         
    12         int result = INT_MAX;
    13         for (int i = 0; i < nums.size(); i++) {
    14             int tempSum = nums[i];
    15             if (tempSum >= s) return 1;
    16             for (int j = i + 1; j < nums.size(); j++) {
    17                 tempSum += nums[j];
    18                 if (tempSum >= s) {
    19                     result = min(result, j - i + 1);
    20                     break;
    21                 }
    22             }
    23         }
    24         return result == INT_MAX ? -1 : result;
    25     }
    26 };
    View Code
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  • 原文地址:https://www.cnblogs.com/amazingzoe/p/5820400.html
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