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338.counting-bits.cpp
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78 lines (74 loc) · 1.56 KB
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// Tag: Dynamic Programming, Bit Manipulation
// Time: O(N)
// Space: O(1)
// Ref: -
// Note: -
// Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's in the binary representation of i.
//
// Example 1:
//
// Input: n = 2
// Output: [0,1,1]
// Explanation:
// 0 --> 0
// 1 --> 1
// 2 --> 10
//
// Example 2:
//
// Input: n = 5
// Output: [0,1,1,2,1,2]
// Explanation:
// 0 --> 0
// 1 --> 1
// 2 --> 10
// 3 --> 11
// 4 --> 100
// 5 --> 101
//
//
// Constraints:
//
// 0 <= n <= 105
//
//
// Follow up:
//
// It is very easy to come up with a solution with a runtime of O(n log n). Can you do it in linear time O(n) and possibly in a single pass?
// Can you do it without using any built-in function (i.e., like __builtin_popcount in C++)?
//
//
class Solution {
public:
vector<int> countBits(int n) {
vector<int> res(n + 1, 0);
for (int i = 1; i <= n; i++) {
int n = i;
while (n > 0) {
res[i] += 1;
n = n & (n - 1);
}
}
return res;
}
};
class Solution {
public:
vector<int> countBits(int n) {
vector<int> dp(n + 1, 0);
for (int i = 1; i <= n; i++) {
dp[i] = dp[i & (i - 1)] + 1;
}
return dp;
}
};
class Solution {
public:
vector<int> countBits(int n) {
vector<int> dp(n + 1, 0);
for (int i = 1; i <= n; i++) {
dp[i] = dp[i >> 1] + (i & 1);
}
return dp;
}
};