209. Minimum Size Subarray Sum

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Problem statement:

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.

Solution:

It looks like 53. Maximum Subarray. But they are different solutions. I solve this problem by two pointers, left and right. The return value is the min length which is greater than or equal to the target. I can control the movement of these two pointers to update the min length.

Basic idea:

• If sum >= target, left moves to the end and minus the value of left, meanwhile update the min length.
• One terminating condition is left > right, that means we already find the min length is 1 and return. Otherwise, I find until the end.
• If sum < target, right moves to the end and plus the value of right.

Since left and right move at most n steps. Time complexity is O(2 * n) ---> O(n).

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        if(nums.empty()){
            return 0;
        }
        int left = 0; 
        int right = 0;
        int sum = nums[left];
        int len = INT_MAX;
        while(left <= right && left < nums.size() && right < nums.size()){
            // if current sum is a solution
            if(sum >= s){
                // update the len
                len = min(len, right - left + 1);
                // update the sum
                // minus first and move left towards end
                sum -= nums[left++];
            } else {
                // the sum is still less than target
                // move right to the end and update the sum
                sum += nums[++right];
            }
        }
        // return 0 if no answer
        return len == INT_MAX ? 0 : len;
    }
};

209. Minimum Size Subarray Sum

标签：font   vector   com   ref   ++   already   nbsp   turn   ready   

原文地址：http://www.cnblogs.com/wdw828/p/6876434.html