leetcode 209. Minimum Size Subarray Sum

最新推荐文章于 2024-06-16 11:54:10 发布

转载最新推荐文章于 2024-06-16 11:54:10 发布 · 100 阅读

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原文链接：http://www.cnblogs.com/exhausttolive/p/10618318.html

本文探讨了使用滑动窗口和前缀和方法解决寻找满足特定条件的最小子数组长度问题。通过两种不同的实现方式，包括直接滑动窗口遍历和结合二分查找优化的前缀和方法，详细解析了算法的设计思路与实现细节。

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using sliding window

class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int l = 0, N = nums.length;
        int ret = N + 1, sum = 0;
        for (int r = 0; r < N; ++r) {
            sum += nums[r];
            for (;l <= r && sum >= s;++l) {
                ret = Math.min(ret, r - l + 1);
                sum -= nums[l];
            }
        }
        return ret == N + 1 ? 0: ret;
    }
}

And try to Using prefix sum method, the rightmost index with value <= current_sum - s must be found.

Since the prefix sum array is monotonically increasing, we can use binary search to find the target index.

But the corner case really takes a while to deal with.

class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int N = nums.length;
        if (N == 0) return 0;
        for (int i = 1; i < N; ++i) nums[i] += nums[i - 1];
        int ret = nums[0] >= s ? 1: N + 1;
        for (int i = 1; i < N; ++i) {
            if (nums[i] < s) continue;
            int target = nums[i] - s;
            if (nums[0] > target) {
                ret = Math.min(ret, i + 1);
                continue;
            }
            int l = 0, r = i - 1;
            while (l < r) {
                int mid = l + (r - l + 1) / 2;
                if (nums[mid] <= target) {
                    l = mid;
                }
                else {
                    r = mid - 1;
                }
            }
            ret = Math.min(ret, i - l);
        }
        return ret == N + 1? 0: ret;
    }
}

Another option is using Arrays.binary search. It makes things a little simpler. But also have to deal with the annoying "what if nums[0] > target" problem.

class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int N = nums.length;
        if (N == 0) return 0;
        for (int i = 1; i < N; ++i) nums[i] += nums[i - 1];
        int ret = nums[0] >= s ? 1: N + 1;
        for (int i = 1; i < N; ++i) {
            if (nums[i] < s) continue;
            int target = nums[i] - s;
            int idx = Arrays.binarySearch(nums, 0, i, target);
            idx = idx < -1? (-idx - 2): idx;//deal with the annoying problem.
            ret = Math.min(ret, i - idx);
        }
        return ret == N + 1? 0: ret;
    }
}

转载于:https://www.cnblogs.com/exhausttolive/p/10618318.html