[LeetCode163]Missing Ranges

Given a sorted integer array where the range of elements are [lower, upper] inclusive, return its missing ranges.

For example, given [0, 1, 3, 50, 75], lower = 0 and upper = 99, return ["2", "4->49", "51->74", "76->99"].

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你还记得summary range 那道题吗？眼熟吗？这道题就是反过来了，乍一看好简单，结果好多edge case cover不了，越写越成了spaghetti code. 因为太笨。

参考大家智慧的结晶吧！最后发现就三类：

case1: lower < num[0]; then we need a [lower, num[0]-1]

case2: check nums[i] == nums[i-1] + 1 ? continue : res.push_back(getRange(nums[i-1]+1, nums[i]-1);

case3: num[n-1] < upper; then we need a[num[n-1]+1, upper];

除此之外， 多写helper function 是人类进步的阶梯。比如我一直要getRange, 那就写个function 吧：

string getRange(int s, int e){
if(s == e) return to_string(s);
else return to_string(s) + "->" + to_string(e);
}

最终code：就很简单啊

class Solution {
public:
vector<string> findMissingRanges(vector<int>& nums, int lower, int upper) {
vector<string> res;
if(nums.empty()){
res.push_back(getRange(lower, upper));
return res;
}
if(lower < nums[0]){
res.push_back(getRange(lower, nums[0]-1));
}
for (int i = 1; i<nums.size(); ++i) {
if (nums[i]-nums[i-1] >1) {
res.push_back(getRange(nums[i-1]+1, nums[i]-1));
}
}
if (upper > nums[nums.size()-1]) {
res.push_back(getRange(nums[nums.size()-1]+1, upper));
}
return res;
}
string getRange(int start, int end){
return (start == end) ? to_string(start) : to_string(start) + "->" + to_string(end);
}
};