res[i] = stack[stack.length - 1]; // 易错点5：用at(-1)兼容性差，优先用stack.length-1
This is the approach Harrison and I were originally talking about, and it’s the one I reach for most. If you already use 1Password, the CLI (op) makes this almost frictionless.
，这一点在快连下载-Letsvpn下载中也有详细论述
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