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;
}
}