//1.单词映射匹配检测
function testMatch( pattern , str ) {
pattern = pattern.split('');
pattern;
str = str.split(' ');
str;
if(!pattern.length && !str.length) return true;
if(pattern.length !== str.length) return false;
var setPattern = Array.from(new Set(pattern));
var setStr = Array.from(new Set(str));
for(let i=0;i<pattern.length;i++){
if(setStr[setPattern.indexOf(pattern[i])]!==str[i]) return false;
}
return true;
}
var pattern = "effe",str="good bad bad bad";
console.log(testMatch(pattern,str));
//给定正整数N，调整顺序获得一个比N大的最小正整数
function findGreaterNum(n) {
if(n<=10) return n;
var list = String(n).split('');
list;
for(var i=list.length-1;i>0;i--){
for(var j=i-1;j>0;j--){
if(list[i]>list[j]){
var temp = list[j];
list[j] = list[i];
list[i] = temp;
return list.join('');
}
}
return -1;
}
}
var n =1234;
console.log(findGreaterNum(n));
//给定正整数数组A和正整数T，在A中找出两个连续子数组，两个子数组不能相交且两个子数组的和为T，返回方案中两个子数组长度和的最小值。
function minSumOfLengths( A , T ) {
var list = [];
for(var i=0;i<A.length;i++){
var sum = T;
list[i]=[];
for(var j=i;j<A.length;j++){
sum-=A[j];
list[i].push(A[j]);
if(sum<0){
list.pop();
break;
}
if(sum===0) break;
}
}
var newlist=[];
for(var k=0;k<list.length;k++){
if(list[k]) newlist.push(list[k]);
}
if(newlist.length<2) return -1;
newlist = newlist.sort((a,b)=>a.length-b.length);
return newlist[0].length+newlist[1].length;
}
var a1=[3,2,2,4,3],t1=3;
var a2=[7,3,4,7],t2=7;
var a3=[5,5,4,4,5],t3=3;
console.log(minSumOfLengths(a3,t1));
//[3,2,2,4,3],3 ==> 2
//[7,3,4,7],7 ==> 2
//[5,5,4,4,5],3 ==>-1 
