【Leetcode_easy】754. Reach a Number
problem
solution1:
class Solution {
public:
int reachNumber(int target) {
target = abs(target);
long n = ceil(0.5*(sqrt(+8.0*target)-1.0));
long sum = n*(n+)/;
if(sum==target) return n;
long d = sum - target;
if((d&) == ) return n;
return n + ((n&) ? : );
}
};
solution2:
class Solution {
public:
int reachNumber(int target) {
target = abs(target);
int res = , sum = ;
while(sum<target || (sum-target)%==)
{
res++;
sum += res;
}
return res;
}
};
solution3:solution1的精简版;
class Solution {
public:
int reachNumber(int target) {
int n = ceil((sqrt(+8.0*abs(target))-)/), d = n*(n+)*0.5-abs(target);
return n + (d%)*(n%+);
}
};
参考
1. Leetcode_easy_754. Reach a Number;
2. Grandyang;
完
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