字符串字符匹配
http://www.nowcoder.com/practice/22fdeb9610ef426f9505e3ab60164c93
while True:
try:
S, T = input(), input()
if set(S) & set(T) == set(S):
print('true')
else:
print('false')
except:
break
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const flattern = (obj) => {
const res = {};
const dfs = (curr, path) => {
if(typeof curr === 'object' && curr !== null) {
const isArray = Array.isArray(curr);
for(let key in curr) {
const newPath = path ? isArray ? `${path}[${key}]` : `${path}.${key}` : key;
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}
} else {
res[path] = curr
}
}
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