滑動窗口(四)——leetcode No.209. Minimum Size Subarray Sum

題目:

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).

分析:

這個題如果只考慮O(n)的解法,就是道easy難度的題目,因爲就變成了典型的滑動窗口,設置兩個變量end,start,每次先滑動窗口的end,直到sum>=s爲止,然後滑動start使得sum小於s爲止,記錄此時end與start差,維護最小長度,繼續滑動窗口直到end到達邊界.

顯然複雜度爲O(n),因爲end變量是一直在動的,而start又受制於end變化,外層僅有end的遍歷.

考慮O(nlogn)的複雜度,由於有logn,一般是用二分法,接下來就是如何二分的問題,我們可以考慮用連續值sum來判斷,代碼寫得很詳細了,可以看代碼.

代碼:

public class Solution{
    public int minSubArrayLen(int s,int [] nums){
        return solveNlogN(s,nums);
    }
    private int solveNlogN(int s,int [] nums){
        int [] sums = new int[nums.length+1];
        // init the sums
        for(int i=1;i<sums.length;i++){
            sums[i]=sums[i-1]+nums[i-1];
        }
        int minLen = Integer.MAX_VALUE;
        //二分查找序列
        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;
            }
            minLen=Math.min(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;
    }
    private int solveN(int s,int [] nums){
        int start=0,end=0,sum=0,minLen=Integer.MAX_VALUE;
        while(end<nums.length){
            while(end<nums.length&&sum<s){
                sum+=nums[end++];
            }
            if(sum<s)
                break;
            while(start<end&&sum>=s){
                sum-=nums[start++];
            }
            minLen=Math.min(minLen,end-start+1);
        }
        return minLen==Integer.MAX_VALUE?0:minLen;
    }

}