思路：倒序单调栈。弹出所有 ≤ 当前身高的元素（这些人都能被看到），count 为弹出数量；若栈非空，还能看到栈顶（第一个更高的人），故 +1。能看到的人数 = count + (栈非空 ? 1 : 0)。
The Beckham bandwagon gives Cruz many advantages, of course. The number of guitars he played on stage on Wednesday would be beyond the reach of the average new artist, for a start.，详情可参考heLLoword翻译官方下载
Add CBS News on Google，更多细节参见搜狗输入法2026
Author Correction: Global subsidence of river deltas