zoukankan      html  css  js  c++  java
  • 209. Minimum Size Subarray Sum

    https://leetcode.com/problems/minimum-size-subarray-sum/#/solutions

    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.
    
    click to show more practice.
    
    More practice:
    If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).

     Since the given array contains only positive integers, the subarray sum can only increase by including more elements. Therefore, you don't have to include more elements once the current subarray already has a sum large enough. This gives the linear time complexity solution by maintaining a minimum window with a two indices.

    窗口指针

    public int minSubArrayLen(int s, int[] a) {
      if (a == null || a.length == 0)
        return 0;
      
      int i = 0, j = 0, sum = 0, min = Integer.MAX_VALUE;
      
      while (j < a.length) {
        sum += a[j++];
        
        while (sum >= s) {
           min = Math.min(min, j - i);
           sum -= a[i++]; } } return min == Integer.MAX_VALUE ? 0 : min; }

    O(NlogN) 

    private int solveNLogN(int s, int[] nums) {
            int[] sums = new int[nums.length + 1];
            for (int i = 1; i < sums.length; i++) sums[i] = sums[i - 1] + nums[i - 1];
            int minLen = Integer.MAX_VALUE;
            for (int i = 0; i < sums.length; i++) {
                int end = binarySearch(i + 1, sums.length - 1, sums[i] + s, sums);
                if (end == sums.length) break;
                if (end - i < minLen) minLen = end - i;
            }
            return minLen == Integer.MAX_VALUE ? 0 : minLen;
        }
        
        private int binarySearch(int lo, int hi, int key, int[] sums) {
            while (lo <= hi) {
               int mid = (lo + hi) / 2;
               if (sums[mid] >= key){
                   hi = mid - 1;
               } else {
                   lo = mid + 1;
               }
            }
            return lo;
        }
    

     为什么我这么做不对?

     public int minSubArrayLen(int s, int[] nums) {
            if (nums == null || nums.length == 0) {
                return 0;
            }
          int[] sums = new int[nums.length + 1];
            for (int i = 1; i < sums.length; i++) sums[i] = sums[i - 1] + nums[i - 1];
            int minLen = Integer.MAX_VALUE;
            for (int i = 0; i < sums.length; i++) {
                int end = Arrays.binarySearch(sums, i + 1, sums.length, sums[i] + s);
                if (end == sums.length) break;
                if (end < 0) {
                    end = -(end + 1);
                }
                if (end - i < minLen) minLen = end - i;
            }
            return minLen == Integer.MAX_VALUE ? 0 : minLen;
        }
    

      

    Input:7 [2,3,1,2,4,3]
    Output:1
    Expected:2

     

  • 相关阅读:
    Java平台调用Python平台已有算法(附源码及解析)
    java截取避免空字符丢失
    Java集合对象比对
    idea中的beautiful插件-自动生成对象set方法
    idea下maven命令打包不同配置
    提纲
    标记语言入门
    react入门
    深入理解React、Redux
    css伪类 伪元素
  • 原文地址:https://www.cnblogs.com/apanda009/p/7095780.html
Copyright © 2011-2022 走看看