338.Counting Bits
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:
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| Input: n = 2
Output: [0,1,1]
Explanation:
0 --> 0
1 --> 1
2 --> 10
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Example 2:
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| Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101
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Brian Kernighan 算法：
时间复杂度：O(nlogn)
空间复杂度：O(1)
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| class Solution {
public:
int countOnes(int x) {
int ones = 0;
while (x > 0) {
x &= (x - 1);
ones++;
}
return ones;
}
vector<int> countBits(int n) {
vector<int> bits(n + 1);
for (int i = 0; i <= n; i++) {
bits[i] = countOnes(i);
}
return bits;
}
};
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动态规划：
时间复杂度：O(n)
空间复杂度：O(1)
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| class Solution {
public:
vector<int> countBits(int n) {
vector<int> bits(n + 1);
int highBit = 0;
for (int i = 1; i <= n; i++) {
if ((i & (i - 1)) == 0) {
highBit = i;
}
bits[i] = bits[i - highBit] + 1;
}
return bits;
}
};
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| class Solution {
public:
vector<int> countBits(int n) {
vector<int> bits(n + 1);
for (int i = 1; i <= n; i++) {
bits[i] = bits[i >> 1] + (i & 1);
}
return bits;
}
};
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| class Solution {
public:
vector<int> countBits(int n) {
vector<int> bits(n + 1);
for (int i = 1; i <= n; i++) {
bits[i] = bits[i & (i - 1)] + 1;
}
return bits;
}
};
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