Leetcode - MissingRange

最新推荐文章于 2024-06-17 19:24:35 发布

最新推荐文章于 2024-06-17 19:24:35 发布 · 139 阅读

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本文介绍了一种用于查找已排序整数数组中缺失区间的高效算法。该算法适用于范围在[0,99]内的整数数组，并能正确处理空数组及包含所有可能元素的情况。通过两个实现方法展示了如何找出数组中的缺失部分并将其格式化为指定形式。

Given a sorted integer array where the range of elements are [0, 99] inclusive, return its missing ranges.
For example, given [0, 1, 3, 50, 75], return [“2”, “4->49”, “51->74”, “76->99”]

[TIPS] 1.特殊情况向面试官确认，数组为空或者包含全部元素返回什么

2. 注意可扩展性，题意为[0,99]，扩展为[start, end]
3 findMissingRanges2方法未在online上跑过，简单自测， findMissingRanges转载自链接以备忘，
优点在于无需额外处理特殊情形

    public List<String> findMissingRanges2(int[] vals, int start, int end) {
        List<String> ranges = new ArrayList<String>();
        if (vals == null || vals.length == 0) {
            ranges.add(start + "->" + end);
            return ranges;
        }
        // int need = 0; // 如果start ！= 0 就错了
        int need = start;
        for (int i = 0; i < vals.length; i++) {
            if (vals[i] > need) {
                ranges.add(getRange(need, vals[i] - 1));
            }
            need = vals[i] + 1;
        }
        if (need <= end)
            ranges.add(getRange(need, end));
        return ranges;
    }

    // 转载：http://www.danielbit.com/blog/puzzle/leetcode/leetcode-missing-ranges
    public List<String> findMissingRanges(int[] vals, int start, int end) {
        List<String> ranges = new ArrayList<String>();
        if (vals == null || vals.length == 0) {
            ranges.add(start + "->" + end);
            return ranges;
        }
        int prev = start - 1;
        for (int i = 0; i <= vals.length; i++) {
            int curr = (i == vals.length) ? end + 1 : vals[i];
            if (curr - prev >= 2) {
                ranges.add(getRange(prev + 1, curr - 1));
            }
            prev = curr;
        }
        return ranges;
    }

    private String getRange(int from, int to) {
        return (from == to) ? String.valueOf(from) : from + "->" + to;
    }