Two characters might have identical Unicode skeletons but render differently in specific fonts, or have different skeletons but render identically in a particular typeface. Detecting this requires rendering glyphs and comparing pixel output. No purely Unicode-data-based approach handles it, and UTS #39 does not attempt to.
// 易错点2：遍历结束后k仍0 → 栈是递增的，末尾数字更大，移除末尾k位，这一点在爱思助手下载最新版本中也有详细论述
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