此题若不考虑极大值极小值相关的corner case是简单的,如base version,当前leetcode的test case没有包含这些边缘case,因此是可以通过的。下面给出两种实现,base version和做了防溢出处理的robust version,并提供了相应的test case。
遇到整数数组问题,需要特别留意溢出问题,面试时可先写出base version然后再考虑增加放溢出处理,不然后者可能严重干扰主干思路,对我来说,添加上防溢出处理的时间远多于写出base version的时间。
public class MissingRanges { @Test public void testMissingRanges() { MissingRanges obj = new MissingRanges(); List<String> expect = new ArrayList<String>(); // case 1 expect.add("-1->2147483646"); int[] nums = {Integer.MAX_VALUE}; List<String> actual1 = obj.findMissingRanges1(nums, -1, Integer.MAX_VALUE); List<String> actual2= obj.findMissingRanges2(nums, -1, Integer.MAX_VALUE); Assert.assertEquals(false, checker(expect, actual1)); Assert.assertEquals(true, checker(expect, actual2)); // case2 expect.clear(); int[] nums2 = {Integer.MAX_VALUE}; List<String> actual3 = obj.findMissingRanges1(nums2, Integer.MAX_VALUE, Integer.MAX_VALUE); List<String> actual4= obj.findMissingRanges2(nums2, Integer.MAX_VALUE, Integer.MAX_VALUE); Assert.assertEquals(false, checker(expect, actual3)); Assert.assertEquals(true, checker(expect, actual4)); } public boolean checker(List<String> expect, List<String> actual) { if (expect.size() != actual.size()) return false; for (int i = 0; i < expect.size(); i++) { if (!expect.get(i).equals(actual.get(i))) return false; } return true; } // more robust version, take care of overflow edge case public List<String> findMissingRanges2(int[] nums, int lower, int upper) { List<String> ret = new ArrayList<String>(); if (nums == null) return ret; for (int i = 0; i < nums.length; i++) { if (lower < nums[i]) { if (lower < nums[i] - 1) {// avoid overflow ret.add("" + lower + "->" + (nums[i] - 1)); } else { ret.add(String.valueOf(lower)); } } if (nums[i] < Integer.MAX_VALUE) // avoid overflow lower = nums[i] + 1; else lower = Integer.MAX_VALUE; } // process lower==upper for case like: [-1], -1, 0 // 添加nums[nums.length -1] != Integer.MAX_VALUE 这个判断,考虑case: // [Integer.MAX_VALUE], Integer.MAX_VALUE, Integer.MAX_VALUE if (lower < upper) { ret.add("" + lower + "->" + upper); } else if (lower == upper && !(nums.length > 0 && nums[nums.length -1] == Integer.MAX_VALUE)) { ret.add(String.valueOf(lower)); } return ret; } // base version public List<String> findMissingRanges1(int[] nums, int lower, int upper) { List<String> ret = new ArrayList<String>(); if (nums == null) return ret; for (int i = 0; i < nums.length; i++) { if (lower < nums[i]) { if (nums[i] - lower > 1) { ret.add("" + lower + "->" + (nums[i] - 1)); } else { ret.add(String.valueOf(lower)); } } lower = nums[i] + 1; } if (lower < upper) { ret.add("" + lower + "->" + upper); } else if (lower == upper) { ret.add(String.valueOf(lower)); } return ret; } }