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29 changes: 29 additions & 0 deletions problems/0053.最大子序和(动态规划).md
Original file line number Diff line number Diff line change
Expand Up @@ -239,6 +239,35 @@ function maxSubArray(nums: number[]): number {
return resMax
}
```
### C:
```c
bool isSubsequence(char *s, char *t)
{
bool **dp = malloc(sizeof(bool *) * (strlen(s) + 1));
for (int i = 0; i <= strlen(s); i++)
{
dp[i] = malloc(sizeof(bool) * (strlen(t) + 1));
}
// 初始化
for (int i = 0; i <= strlen(s); i++)
{
dp[i][0] = false;
}
for (int j = 0; j <= strlen(t); j++)
{
dp[0][j] = true;
}
// 遍历
for (int i = 1; i <= strlen(s); i++)
{
for (int j = 1; j <= strlen(t); j++)
{
dp[i][j] = s[i - 1] == t[j - 1] ? dp[i - 1][j - 1] : dp[i][j - 1];
}
}
return dp[strlen(s)][strlen(t)];
}
```



39 changes: 39 additions & 0 deletions problems/0115.不同的子序列.md
Original file line number Diff line number Diff line change
Expand Up @@ -371,6 +371,45 @@ impl Solution {
}
}
```
### C:
```c
#define MOD 1000000007
int numDistinct(char *s, char *t)
{
//dp[i][j]: t前i个字符,s前j个字符,s子序列中t出现的个数
int **dp = malloc(sizeof(int *) * (strlen(t) + 1));
for (int i = 0; i <= strlen(t); i++)
{
dp[i] = malloc(sizeof(int) * (strlen(s) + 1));
}

// 初始化
for (int i = 0; i <= strlen(t); i++)
{
dp[i][0] = 0;
}
for (int j = 0; j <= strlen(s); j++)
{
dp[0][j] = 1;
}

// 遍历
for (int i = 1; i <= strlen(t); i++)
{
for (int j = 1; j <= strlen(s); j++)
{
if (s[j - 1] == t[i - 1])
{
dp[i][j] = dp[i - 1][j - 1] + dp[i][j - 1];
}
else
{
dp[i][j] = dp[i][j - 1];
}
}
}
return dp[strlen(t)][strlen(s)] % MOD;
}
```


37 changes: 36 additions & 1 deletion problems/0435.无重叠区间.md
Original file line number Diff line number Diff line change
Expand Up @@ -440,7 +440,7 @@ impl Solution {
}
```
### C

- 按右边界排序
```c
// 按照区间右边界排序
int cmp(const void * var1, const void * var2){
Expand All @@ -465,6 +465,41 @@ int eraseOverlapIntervals(int** intervals, int intervalsSize, int* intervalsColS
return intervalsSize - count;
}
```
- 按左边界排序
```c
// 按照区间左边界排序
int cmp(const void *a, const void *b)
{
return *((int **)a)[0] - *((int **)b)[0];
}
int eraseOverlapIntervals(int **intervals, int intervalsSize, int *intervalsColSize)
{
if (intervalsSize == 0)
{
return 0;
}
qsort(intervals, intervalsSize, sizeof(intervals[0]), cmp);
// count: 重叠的区间数量
int count = 0;
// end: 区间分割点
int end = intervals[0][1];
for (int i = 1; i < intervalsSize; i++)
{
if (intervals[i][0] < end)
{
// 说明有重合, end取2个区间中较小的右边界
end = fmin(end, intervals[i][1]);
count++;
}
else
{
// 没有重合,更新end
end = intervals[i][1];
}
}
return count;
}
```



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