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Commit 15d6c5f

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Improvement
Signed-off-by: begeekmyfriend <begeekmyfriend@gmail.com>
1 parent 4fab7e0 commit 15d6c5f

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8 files changed

+40
-25
lines changed

8 files changed

+40
-25
lines changed

‎0039_combination_sum/combination_sum.cc‎

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -20,7 +20,7 @@ class Solution {
2020
} else {
2121
for (int i = start; i < candidates.size(); i++) {
2222
stack.push_back(candidates[i]);
23-
/* The elements in solution can be duplicate for the purpose of the problem */
23+
/* The elements in solution can be taken as many times as you can for the purpose of the problem */
2424
dfs(candidates, i, target - candidates[i], res);
2525
stack.pop_back();
2626
}

‎0045_jump_game_ii/jump_game.c‎

Lines changed: 4 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -12,10 +12,10 @@ static int jump(int* nums, int numsSize)
1212
int i, right = 0;
1313
int steps = 0;
1414
int fartest = 0;
15-
/* 1. Exhaust all the right boundries in the location range of [i...right]
16-
* 2. When the search ends up with i==right, update the right boundry as
17-
* the fartest position.
18-
* 3. When the search ends up with i==right, it records as one jump step */
15+
/* 1. Exhaust all the right boundries in the location range of [i...farthest]
16+
* 2. When i reaches the farthest boundary, update the farthest boundry
17+
* and the step number.
18+
* 3. Apply condition i < size - 1 and iterator i++ to avoid overflow. */
1919
for (i = 0; i < numsSize; i++) {
2020
fartest = max(i + nums[i], fartest);
2121
if (i == right) {

‎0045_jump_game_ii/jump_game.cc‎

Lines changed: 7 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -8,14 +8,14 @@ class Solution {
88
int steps = 0;
99
int right = 0;
1010
int farthest = 0;
11-
// 1. Exhaust all the right boundries in the location range of [i...right]
12-
// 2. When the search ends up with i==right, update the right boundry as
13-
// the fartest position.
14-
// 3. When the search ends up with i==right, it records as one jump step */
11+
// 1. Exhaust all the right boundries in the location range of [i...farthest]
12+
// 2. When i reaches the farthest boundary, update the farthest boundry
13+
// and the step number.
14+
// 3. Apply condition i < size - 1 and iterator i++ to avoid overflow.
1515
for (int i = 0; i < nums.size() - 1; i++) {
16-
fartest = max(i + nums[i], fartest);
17-
for (i == right) {
18-
right = fartest;
16+
right = max(i + nums[i], right);
17+
if (i == farthest) {
18+
farthest = right;
1919
steps++;
2020
}
2121
}

‎0069_sqrt/sqrt.c‎

Lines changed: 3 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -60,6 +60,9 @@ int mySqrt(int x)
6060
unsigned int lo = 1;
6161
unsigned int hi = (unsigned int) x;
6262
unsigned int mid = lo + (hi - lo) / 2;
63+
// Firstly test mid > x / mid to decide whether hi = mid;
64+
// else then test mid + 1 > x / (mid + 1) to decide whether the mid is located;
65+
// Otherwise assign low = mid.
6366
for (; ;) {
6467
if (mid > x/mid) {
6568
hi = mid;

‎0069_sqrt/sqrt.cc‎

Lines changed: 3 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -11,6 +11,9 @@ class Solution {
1111

1212
unsigned int lo = 1, hi = x;
1313
unsigned int mid = (lo + hi) / 2;
14+
// Firstly test mid > x / mid to decide whether hi = mid;
15+
// else then test mid + 1 > x / (mid + 1) to decide whether the mid is located;
16+
// Otherwise assign low = mid.
1417
for (; ;) {
1518
if (mid > x / mid) {
1619
hi = mid;

‎0076_minimum_window_substring/window_substring.cc‎

Lines changed: 10 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -11,20 +11,25 @@ class Solution {
1111
}
1212

1313
int l = 0, r = 0;
14-
int need_to_meet = t.length();
15-
int start, min_len = INT_MAX;
14+
int hit_num = 0;
15+
int start = 0, min_len = INT_MAX;
1616
while (r < s.length()) {
17+
// counting each letter in the string. The zero and positive
18+
// countings indicate ones in pattern. And the negative ones
19+
// indicate those out of the pattern.
1720
if (--count[s[r++]] >= 0) {
18-
need_to_meet--;
21+
hit_num++;
1922
}
2023

21-
while (need_to_meet == 0) {
24+
while (hit_num == t.length()) {
2225
if (r - l < min_len) {
2326
start = l;
2427
min_len = r - l;
2528
}
29+
// The countings of the letter larger than zero shall be
30+
// the ones in the pattern.
2631
if (++count[s[l++]] > 0) {
27-
need_to_meet++;
32+
hit_num--;
2833
}
2934
}
3035
}

‎0137_single_number_ii/single_number.c‎

Lines changed: 1 addition & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -39,6 +39,7 @@ static int singleNumber(int *nums, int numsSize)
3939
count[i]++;
4040
}
4141
}
42+
/* The specified bit counting should be multiple of 3 without the outlier */
4243
mask |= (count[i] % 3) << i;
4344
}
4445
return mask;

‎0190_reverse_bits/reverse_bits.c‎

Lines changed: 11 additions & 8 deletions
Original file line numberDiff line numberDiff line change
@@ -4,14 +4,17 @@
44

55
static uint32_t reverseBits(uint32_t n)
66
{
7-
int i;
8-
uint32_t res = 0;
9-
for (i = 0; i < 32; i++) {
10-
res <<= 1;
11-
res |= n & 0x1;
12-
n >>= 1;
13-
}
14-
return res;
7+
const uint32_t MASK1 = 0x55555555;
8+
const uint32_t MASK2 = 0x33333333;
9+
const uint32_t MASK4 = 0x0f0f0f0f;
10+
const uint32_t MASK8 = 0x00ff00ff;
11+
12+
// Extract and swap the even and odd bit groups.
13+
n = (n & MASK1) << 1 | ((n >> 1) & MASK1);
14+
n = (n & MASK2) << 2 | ((n >> 2) & MASK2);
15+
n = (n & MASK4) << 4 | ((n >> 4) & MASK4);
16+
n = (n & MASK8) << 8 | ((n >> 8) & MASK8);
17+
return n << 16 | n >> 16;
1518
}
1619

1720
int main(int argc, char **argv)

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