You are given an array nums of non-negative integers. nums is considered special if there exists a number x such that there are exactly x numbers in nums that are greater than or equal to x.
Notice that x does not have to be an element in nums.
Return x if the array is special, otherwise, return -1. It can be proven that if nums is special, the value for x is unique.
Example 1:
Input: nums = [3,5]
Output: 2
Explanation: There are 2 values (3 and 5) that are greater than or equal to 2.
Example 2:
Input: nums = [0,0]
Output: -1
Explanation: No numbers fit the criteria for x.
If x = 0, there should be 0 numbers >= x, but there are 2.
If x = 1, there should be 1 number >= x, but there are 0.
If x = 2, there should be 2 numbers >= x, but there are 0.
x cannot be greater since there are only 2 numbers in nums.
Example 3:
Input: nums = [0,4,3,0,4]
Output: 3
Explanation: There are 3 values that are greater than or equal to 3.
Example 4:
Input: nums = [3,6,7,7,0]
Output: -1
Constraints:
1 <= nums.length <= 100
0 <= nums[i] <= 1000
这题有点儿不好理解。我这里把数字进行了分组,然后判断比该组元素大的元素个数是否在上组的数字和该组数字的开闭范围内。
class SpecialArrayWithXElementsGreaterThanOrEqualX : public Solution {
public:
void Exec() {
}
int specialArray(vector<int>& nums) {
map<int, int> numsCounter;
int greaterCount = nums.size();
for (auto v : nums) {
auto it = numsCounter.find(v);
if (it == numsCounter.end()) {
numsCounter.insert(std::make_pair(v, 1));
} else {
it->second++;
}
}
int prevRange = -1;
while (!numsCounter.empty()) {
auto it = numsCounter.begin();
if (greaterCount <= prevRange) {
return -1;
}
if (greaterCount <= it->first) {
return greaterCount;
}
prevRange = it->first;
greaterCount -= it->second;
numsCounter.erase(it);
}
return -1;
}
};