LeetCode-209-Minimum Size Subarray Sum

Minimum Size Subarray Sum

 

来自 <https://leetcode.com/problems/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 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.

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).

Credits:

Special thanks to @Freezen for adding this problem and creating all test cases.

题目解读：

给定一个长度为n的正整数数组和和一个正整数，找出一个长度最短的子数组，使子数组元素之和sum ≥ s。如果这样的子数组不存在，则返回0.

例如：给定的数组为[2,3,1,2,4,3]和 s=7。则子数组[4,3]的长度最小。

解析：

解法一：从前向后遍历数组，设置两个指针i,j ，在开始时两个指针指向相同的位置。遍历数组时，一个指针i首先保持不动，另外一个指针j向后移动并记录并记录指针i和j之间数组元素的和sum，如果sum<s，则j继续向后移动。如果sum>=s。则记录length=j-i，看其是否小于minlength, 如果小于，则minlength=length；否则minlength保持不变。此时i和j再同时指向i的下一个位置，依次循环。主要利用滑动窗口的思想。

 图片摘自：http://blog.csdn.net/lisonglisonglisong/article/details/45666975

解法二：

摘自http://www.cnblogs.com/tonyluis/p/4564076.html

在方法一中，保持前面的指针不变，让后边的指针移动。在方法二中，维护一个左边界，让右边的正常移动。

解法三：分治法

摘自http://www.huhaoyu.com/leetcode-minimum-size-subarray-sum/

 

解法一代码：

Public int minSubArrayLen(int s, int[] nums) {
		int sum = 0;
		int minlength = nums.length;
		int length = 0;
		for(int i=0; i<nums.length; i++) {
			sum += nums[i];
			int j=0;
			for(j=i+1; j<nums.length; j++ ) {
				if(sum >= s) {
					length = j-i;
					break;
					
				} else {
					sum += nums[j];
				}
			}
			if(j==nums.length && sum >= s) {
				length = j-i;
			}
			if(minlength > length) {
				minlength = length;
			}
			sum = 0;
		}
		if(minlength != nums.length) {
			return minlength;
		} else {
			return 0;
		}
    }

 

 

解法一性能：

 

 

解法二代码：

    public int minSubArrayLen(int s, int[] nums) {
        	int i=0;
		int left = 0;
		int minLength = nums.length;
		int sum = 0;
		int length = 0;
		for(i=0; i<nums.length; i++) {
			sum += nums[i];
			while(sum >= s) {
			    length = i-left+1;
			    //minLength = Math.min(minLength, i-left+1);
			    if(minLength > length) {
			        minLength = length;
			    }
				sum -= nums[left++];
			}
		}
		
		//return minLength==nums.length?0:minLength;
		if(minLength == nums.length) {
		    return 0;
		} else {
		    return minLength;
		}
    }

 

 

解法二性能：

 

 

解法三代码:(仅作参考，提交不成功，但是个很好的思想)

class Solution {
public:
    int MAX_INT = 999999999;
    int minSubArrayLen(int s, vector<int>& nums) {
        if (!nums.size()) return 0;
        return findMinLen(s, nums, 0, nums.size() - 1);
    }
    int findMinLen(int s, vector<int>& nums, int bottom, int top) {
        if (top == bottom) return nums[top] >= s ? 1 : 0;
        int mid = (bottom + top) / 2;
        int left = mid, right = mid + 1, sum = 0, len;
        while (sum < s && (right <= top || left >= bottom)) {
            if  (right > top) {
                sum += nums[left]; --left;
            }
            else if (left < bottom) {
                sum += nums[right]; ++right;
            }
            else if (nums[left] > nums[right]) {
                sum += nums[left]; --left;
            }
            else {
                sum += nums[right]; ++right;
            }
        }
        if (sum >= s) {
            len = right - left - 1;
            int leftLen = findMinLen(s, nums, bottom, mid);
            int rightLen = findMinLen(s, nums, mid + 1, top);
            return minValue(leftLen, rightLen, len);
        }
        else {
            return 0;
        }
    }
    int minValue(int x, int y, int z) {
        if (!x) x = MAX_INT;
        if (!y) y = MAX_INT;
        if (x <= y && x <= z) return x;
        if (y <= x && y <= z) return y;
        return z;
    }
};

 来自 <http://www.huhaoyu.com/leetcode-minimum-size-subarray-sum/>