LeetCode 209. Minimum Size Subarray Sum

问题描述

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.

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

问题分析

给定一个数组nums[],给定一个数target。求nums中连续子集的和等于target。求出这些子集中长度最小的。遇到连续子集算法的实现有两种，一种是滑动窗口，窗口的条件是sum>= target,这种的复杂度是O(n)。第二种是求前n个的和，这里需要把sum[0]设置为0，这个算法的复杂度是O(nlogn)

代码实现

 public int minSubArrayLen(int s, int[] nums) {        //找出和大于s的 nums中的子序列最小的长度        //使用滑动窗口中和》= nums的的窗口,找到        int minLength = Integer.MAX_VALUE;        int leftIndex = 0;        int rightIndex = 0;        int sum = 0;        int length = 0;        while (sum >= s || rightIndex < nums.length) {//如何确定边界            if (sum < s) {// 在小于的时候                length++;                sum = sum + nums[rightIndex];                rightIndex++;            } else {//大于等于的时候                minLength = Math.min(length, minLength);                sum = sum - nums[leftIndex];                leftIndex++;                length--;            }        }        if (minLength == Integer.MAX_VALUE) {            return 0;        }        return minLength;    }

滑动窗口实现。

代码实现二

public int minSubArrayLen(int s, int[] nums) {        if (nums == null || nums.length == 0) {            return 0;        }        int minLenget = Integer.MAX_VALUE;        int[] sums = new int[nums.length + 1];        sums[0] = 0;        for (int i = 1; i < sums.length; i++) {            sums[i] = sums[i - 1] + nums[i - 1];        }        for (int i = 0; i < sums.length; i++) {            int sum = sums[i] + s;            int nexNum = nextIndex(sums, sum, i);            if (nexNum == sums.length) {                break;            }            minLenget = Math.min(minLenget, nexNum - i);        }        if (minLenget == Integer.MAX_VALUE) {            return 0;        }        return minLenget;    }    /**     * 找出sums中第一个等于大于num的值的下标     *     * @param sums     * @param num     * @return     */    protected int nextIndex(int[] sums, int num, int left) {        int start = left;        int end = sums.length - 1;        while (start <= end) {            int mid = (start + end) / 2;            if (sums[mid] == num) {                return mid;            } else if (sums[mid] > num) {                end = mid - 1;            } else if (sums[mid] < num) {                start = mid + 1;            }        }        return start;    }