Double Pointers#

Check if `p` is a subsequence of `s`#

Check whether p is a subsequence of s after some characters of s have been removed.

✅ 最优雅（推荐）——只写一个 if#

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j = 0
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for i in range(m):
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    if not removed[i] and j < len(p) and s[i] == p[j]:
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        j += 1
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return j == len(p)

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int j = 0;
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for (int i = 0; i < m; i++) {
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    if (!removed[i] && j < (int)p.size() && s[i] == p[j]) {
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        j++;
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    }
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}
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return j == (int)p.size();

✅ 更“清爽”的 continue 版（逻辑最直观）#

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j = 0
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for i in range(m):
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    if removed[i]:
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        continue
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    if j < len(p) and s[i] == p[j]:
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        j += 1
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return j == len(p)

(int) 是为了避免 int 和 size_t(无符号) 混着比较导致警告/潜在bug。在 C++ 里如果你把 j 定义成 size_t（无符号），然后写了 j--，会出现一个非常坑的现象：不会变成 -1，而是“下溢”变成一个超级大的数。

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int j = 0;
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for (int i = 0; i < m; i++) {
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    if (removed[i]) continue;
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    if (j < (int)p.size() && s[i] == p[j]) {
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        j++;
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    }
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}
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return j == (int)p.size();

✅ 用 for ch in s 更优雅（但要处理 removed）#

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j = 0
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for ch in s:
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    if j < len(p) and ch == p[j]:
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        j += 1
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return j == len(p)

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int j = 0;
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for (char ch : s) {
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    if (j < (int)p.size() && ch == p[j]) {
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        j++;
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    }
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}
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return j == (int)p.size();

Double Pointers#

Check if p is a subsequence of s#

✅ 最优雅（推荐）——只写一个 if#

✅ 更“清爽”的 continue 版（逻辑最直观）#

✅ 用 for ch in s 更优雅（但要处理 removed）#

Check if `p` is a subsequence of `s`#