Minimum Size Subarray Sum

最新推荐文章于 2024-06-25 02:17:10 发布

转载最新推荐文章于 2024-06-25 02:17:10 发布 · 43 阅读

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原文链接：http://www.cnblogs.com/zengzy/p/5006540.html

本文介绍了解决寻找给定数组中满足指定和条件的最小子数组长度的问题，提供了两种解决方案，一种时间复杂度为O(n)，另一种为O(n log n)。

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 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).

注意：求的是连续子数组

1.o(n)

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int numsSize = nums.size();
        int i=-1,j=0;
        int sum=0;
        int minLen = numsSize+1;
        while(i<j){
            if(sum<s){
                if(j>=numsSize){
                    break;
                }
                sum+=nums[j++];
            }else if(sum>=s){
                minLen = min(minLen,j-i-1);
                sum-=nums[++i];
            }
        }
        return minLen==numsSize+1 ? 0:minLen;
    }
};

2.o(nlgn)

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int numsSize = nums.size();
        int low = 0,high=numsSize+1;
        while(low<high){
            int mid = low+(high-low)/2;
            int sum = 0;
            for(int i=0;i<numsSize;i++){
                sum+=nums[i];
                if(i>=mid){
                    sum-=nums[i-mid];
                }
                if(sum>=s){
                    break;
                }
            }
            if(sum>=s){
                high = mid;
            }else{
                low = mid+1;
            }
        }
        return low==numsSize+1 ? 0:low;
    }
};