找不同使用 javascript
find the difference use javascript
这是一道 leetcode 题。
给定两个仅由小写字母组成的字符串 s 和 t。
字符串t是由字符串s随机打乱后在随机位置再加一个字母生成的
找到t中添加的字母。
https://leetcode.com/problems/find-the-difference/
我正在尝试用散列 table 的方式思考,但似乎我错了。
var findTheDifference = function(s, t) {
var hashTable = {};
var array = s.split('');
array.forEach(function (element) {
hashTable[element] = element;
});
for( var i = 0; i < t.length; i++) {
if(!hashTable.hasOwnProperty(t.charAt(i))) {
return t.charAt(i);
}
}
};
这个错误的代码可以通过示例输入:
Input:
s = "abcd"
t = "abcde"
Output:
e
但对于大字符串会出错
Input :"ymbgaraibkfmvocpizdydugvalagaivdbfsfbepeyccqfepzvtpyxtbadkhmwmoswrcxnargtlswqemafandgkmydtimuzvjwxvlfwlhvkrgcsithaqlcvrihrwqkpjdhgfgreqoxzfvhjzojhghfwbvpfzectwwhexthbsndovxejsntmjihchaotbgcysfdaojkjldprwyrnischrgmtvjcorypvopfmegizfkvudubnejzfqffvgdoxohuinkyygbdzmshvyqyhsozwvlhevfepdvafgkqpkmcsikfyxczcovrmwqxxbnhfzcjjcpgzjjfateajnnvlbwhyppdleahgaypxidkpwmfqwqyofwdqgxhjaxvyrzupfwesmxbjszolgwqvfiozofncbohduqgiswuiyddmwlwubetyaummenkdfptjczxemryuotrrymrfdxtrebpbjtpnuhsbnovhectpjhfhahbqrfbyxggobsweefcwxpqsspyssrmdhuelkkvyjxswjwofngpwfxvknkjviiavorwyfzlnktmfwxkvwkrwdcxjfzikdyswsuxegmhtnxjraqrdchaauazfhtklxsksbhwgjphgbasfnlwqwukprgvihntsyymdrfovaszjywuqygpvjtvlsvvqbvzsmgweiayhlubnbsitvfxawhfmfiatxvqrcwjshvovxknnxnyyfexqycrlyksderlqarqhkxyaqwlwoqcribumrqjtelhwdvaiysgjlvksrfvjlcaiwrirtkkxbwgicyhvakxgdjwnwmubkiazdjkfmotglclqndqjxethoutvjchjbkoasnnfbgrnycucfpeovruguzumgmgddqwjgdvaujhyqsqtoexmnfuluaqbxoofvotvfoiexbnprrxptchmlctzgqtkivsilwgwgvpidpvasurraqfkcmxhdapjrlrnkbklwkrvoaziznlpor"
"qhxepbshlrhoecdaodgpousbzfcqjxulatciapuftffahhlmxbufgjuxstfjvljybfxnenlacmjqoymvamphpxnolwijwcecgwbcjhgdybfffwoygikvoecdggplfohemfypxfsvdrseyhmvkoovxhdvoavsqqbrsqrkqhbtmgwaurgisloqjixfwfvwtszcxwktkwesaxsmhsvlitegrlzkvfqoiiwxbzskzoewbkxtphapavbyvhzvgrrfriddnsrftfowhdanvhjvurhljmpxvpddxmzfgwwpkjrfgqptrmumoemhfpojnxzwlrxkcafvbhlwrapubhveattfifsmiounhqusvhywnxhwrgamgnesxmzliyzisqrwvkiyderyotxhwspqrrkeczjysfujvovsfcfouykcqyjoobfdgnlswfzjmyucaxuaslzwfnetekymrwbvponiaojdqnbmboldvvitamntwnyaeppjaohwkrisrlrgwcjqqgxeqerjrbapfzurcwxhcwzugcgnirkkrxdthtbmdqgvqxilllrsbwjhwqszrjtzyetwubdrlyakzxcveufvhqugyawvkivwonvmrgnchkzdysngqdibhkyboyftxcvvjoggecjsajbuqkjjxfvynrjsnvtfvgpgveycxidhhfauvjovmnbqgoxsafknluyimkczykwdgvqwlvvgdmufxdypwnajkncoynqticfetcdafvtqszuwfmrdggifokwmkgzuxnhncmnsstffqpqbplypapctctfhqpihavligbrutxmmygiyaklqtakdidvnvrjfteazeqmbgklrgrorudayokxptswwkcircwuhcavhdparjfkjypkyxhbgwxbkvpvrtzjaetahmxevmkhdfyidhrdeejapfbafwmdqjqszwnwzgclitdhlnkaiyldwkwwzvhyorgbysyjbxsspnjdewjxbhpsvj"
感谢任何建议~~~~
您可以使用散列 table 进行计数。为原始字符串中的每个字母加一,为处理后的字符串中的每个字母减一。然后只取非空计数的哈希值。
var data1 = "ymbgaraibkfmvocpizdydugvalagaivdbfsfbepeyccqfepzvtpyxtbadkhmwmoswrcxnargtlswqemafandgkmydtimuzvjwxvlfwlhvkrgcsithaqlcvrihrwqkpjdhgfgreqoxzfvhjzojhghfwbvpfzectwwhexthbsndovxejsntmjihchaotbgcysfdaojkjldprwyrnischrgmtvjcorypvopfmegizfkvudubnejzfqffvgdoxohuinkyygbdzmshvyqyhsozwvlhevfepdvafgkqpkmcsikfyxczcovrmwqxxbnhfzcjjcpgzjjfateajnnvlbwhyppdleahgaypxidkpwmfqwqyofwdqgxhjaxvyrzupfwesmxbjszolgwqvfiozofncbohduqgiswuiyddmwlwubetyaummenkdfptjczxemryuotrrymrfdxtrebpbjtpnuhsbnovhectpjhfhahbqrfbyxggobsweefcwxpqsspyssrmdhuelkkvyjxswjwofngpwfxvknkjviiavorwyfzlnktmfwxkvwkrwdcxjfzikdyswsuxegmhtnxjraqrdchaauazfhtklxsksbhwgjphgbasfnlwqwukprgvihntsyymdrfovaszjywuqygpvjtvlsvvqbvzsmgweiayhlubnbsitvfxawhfmfiatxvqrcwjshvovxknnxnyyfexqycrlyksderlqarqhkxyaqwlwoqcribumrqjtelhwdvaiysgjlvksrfvjlcaiwrirtkkxbwgicyhvakxgdjwnwmubkiazdjkfmotglclqndqjxethoutvjchjbkoasnnfbgrnycucfpeovruguzumgmgddqwjgdvaujhyqsqtoexmnfuluaqbxoofvotvfoiexbnprrxptchmlctzgqtkivsilwgwgvpidpvasurraqfkcmxhdapjrlrnkbklwkrvoaziznlpor",
data2 = "qhxepbshlrhoecdaodgpousbzfcqjxulatciapuftffahhlmxbufgjuxstfjvljybfxnenlacmjqoymvamphpxnolwijwcecgwbcjhgdybfffwoygikvoecdggplfohemfypxfsvdrseyhmvkoovxhdvoavsqqbrsqrkqhbtmgwaurgisloqjixfwfvwtszcxwktkwesaxsmhsvlitegrlzkvfqoiiwxbzskzoewbkxtphapavbyvhzvgrrfriddnsrftfowhdanvhjvurhljmpxvpddxmzfgwwpkjrfgqptrmumoemhfpojnxzwlrxkcafvbhlwrapubhveattfifsmiounhqusvhywnxhwrgamgnesxmzliyzisqrwvkiyderyotxhwspqrrkeczjysfujvovsfcfouykcqyjoobfdgnlswfzjmyucaxuaslzwfnetekymrwbvponiaojdqnbmboldvvitamntwnyaeppjaohwkrisrlrgwcjqqgxeqerjrbapfzurcwxhcwzugcgnirkkrxdthtbmdqgvqxilllrsbwjhwqszrjtzyetwubdrlyakzxcveufvhqugyawvkivwonvmrgnchkzdysngqdibhkyboyftxcvvjoggecjsajbuqkjjxfvynrjsnvtfvgpgveycxidhhfauvjovmnbqgoxsafknluyimkczykwdgvqwlvvgdmufxdypwnajkncoynqticfetcdafvtqszuwfmrdggifokwmkgzuxnhncmnsstffqpqbplypapctctfhqpihavligbrutxmmygiyaklqtakdidvnvrjfteazeqmbgklrgrorudayokxptswwkcircwuhcavhdparjfkjypkyxhbgwxbkvpvrtzjaetahmxevmkhdfyidhrdeejapfbafwmdqjqszwnwzgclitdhlnkaiyldwkwwzvhyorgbysyjbxsspnjdewjxbhpsvj",
hash = Object.create(null),
count = function (a) { hash[a] = (hash[a] || 0) + this; },
result;
data1.split('').forEach(count, 1);
data2.split('').forEach(count, -1);
result = Object.keys(hash).filter(function (k) { return hash[k]; });
console.log(result);
一种方法是对字符串进行排序,然后找出它们之间的第一个区别是:
var str = "1122334477422";
var str2 = "15122332247744";
var sorted = str.split('').sort();
var sorted2 = str2.split('').sort()
for(var i = 0; i< sorted2.length; i++){
if(i == sorted2.length || sorted[i] != sorted2[i]){
alert(sorted2[i]);
break;
}
}
这是一道 leetcode 题。
给定两个仅由小写字母组成的字符串 s 和 t。
字符串t是由字符串s随机打乱后在随机位置再加一个字母生成的
找到t中添加的字母。
https://leetcode.com/problems/find-the-difference/
我正在尝试用散列 table 的方式思考,但似乎我错了。
var findTheDifference = function(s, t) {
var hashTable = {};
var array = s.split('');
array.forEach(function (element) {
hashTable[element] = element;
});
for( var i = 0; i < t.length; i++) {
if(!hashTable.hasOwnProperty(t.charAt(i))) {
return t.charAt(i);
}
}
};
这个错误的代码可以通过示例输入:
Input:
s = "abcd"
t = "abcde"
Output:
e
但对于大字符串会出错
Input :"ymbgaraibkfmvocpizdydugvalagaivdbfsfbepeyccqfepzvtpyxtbadkhmwmoswrcxnargtlswqemafandgkmydtimuzvjwxvlfwlhvkrgcsithaqlcvrihrwqkpjdhgfgreqoxzfvhjzojhghfwbvpfzectwwhexthbsndovxejsntmjihchaotbgcysfdaojkjldprwyrnischrgmtvjcorypvopfmegizfkvudubnejzfqffvgdoxohuinkyygbdzmshvyqyhsozwvlhevfepdvafgkqpkmcsikfyxczcovrmwqxxbnhfzcjjcpgzjjfateajnnvlbwhyppdleahgaypxidkpwmfqwqyofwdqgxhjaxvyrzupfwesmxbjszolgwqvfiozofncbohduqgiswuiyddmwlwubetyaummenkdfptjczxemryuotrrymrfdxtrebpbjtpnuhsbnovhectpjhfhahbqrfbyxggobsweefcwxpqsspyssrmdhuelkkvyjxswjwofngpwfxvknkjviiavorwyfzlnktmfwxkvwkrwdcxjfzikdyswsuxegmhtnxjraqrdchaauazfhtklxsksbhwgjphgbasfnlwqwukprgvihntsyymdrfovaszjywuqygpvjtvlsvvqbvzsmgweiayhlubnbsitvfxawhfmfiatxvqrcwjshvovxknnxnyyfexqycrlyksderlqarqhkxyaqwlwoqcribumrqjtelhwdvaiysgjlvksrfvjlcaiwrirtkkxbwgicyhvakxgdjwnwmubkiazdjkfmotglclqndqjxethoutvjchjbkoasnnfbgrnycucfpeovruguzumgmgddqwjgdvaujhyqsqtoexmnfuluaqbxoofvotvfoiexbnprrxptchmlctzgqtkivsilwgwgvpidpvasurraqfkcmxhdapjrlrnkbklwkrvoaziznlpor"
"qhxepbshlrhoecdaodgpousbzfcqjxulatciapuftffahhlmxbufgjuxstfjvljybfxnenlacmjqoymvamphpxnolwijwcecgwbcjhgdybfffwoygikvoecdggplfohemfypxfsvdrseyhmvkoovxhdvoavsqqbrsqrkqhbtmgwaurgisloqjixfwfvwtszcxwktkwesaxsmhsvlitegrlzkvfqoiiwxbzskzoewbkxtphapavbyvhzvgrrfriddnsrftfowhdanvhjvurhljmpxvpddxmzfgwwpkjrfgqptrmumoemhfpojnxzwlrxkcafvbhlwrapubhveattfifsmiounhqusvhywnxhwrgamgnesxmzliyzisqrwvkiyderyotxhwspqrrkeczjysfujvovsfcfouykcqyjoobfdgnlswfzjmyucaxuaslzwfnetekymrwbvponiaojdqnbmboldvvitamntwnyaeppjaohwkrisrlrgwcjqqgxeqerjrbapfzurcwxhcwzugcgnirkkrxdthtbmdqgvqxilllrsbwjhwqszrjtzyetwubdrlyakzxcveufvhqugyawvkivwonvmrgnchkzdysngqdibhkyboyftxcvvjoggecjsajbuqkjjxfvynrjsnvtfvgpgveycxidhhfauvjovmnbqgoxsafknluyimkczykwdgvqwlvvgdmufxdypwnajkncoynqticfetcdafvtqszuwfmrdggifokwmkgzuxnhncmnsstffqpqbplypapctctfhqpihavligbrutxmmygiyaklqtakdidvnvrjfteazeqmbgklrgrorudayokxptswwkcircwuhcavhdparjfkjypkyxhbgwxbkvpvrtzjaetahmxevmkhdfyidhrdeejapfbafwmdqjqszwnwzgclitdhlnkaiyldwkwwzvhyorgbysyjbxsspnjdewjxbhpsvj"
感谢任何建议~~~~
您可以使用散列 table 进行计数。为原始字符串中的每个字母加一,为处理后的字符串中的每个字母减一。然后只取非空计数的哈希值。
var data1 = "ymbgaraibkfmvocpizdydugvalagaivdbfsfbepeyccqfepzvtpyxtbadkhmwmoswrcxnargtlswqemafandgkmydtimuzvjwxvlfwlhvkrgcsithaqlcvrihrwqkpjdhgfgreqoxzfvhjzojhghfwbvpfzectwwhexthbsndovxejsntmjihchaotbgcysfdaojkjldprwyrnischrgmtvjcorypvopfmegizfkvudubnejzfqffvgdoxohuinkyygbdzmshvyqyhsozwvlhevfepdvafgkqpkmcsikfyxczcovrmwqxxbnhfzcjjcpgzjjfateajnnvlbwhyppdleahgaypxidkpwmfqwqyofwdqgxhjaxvyrzupfwesmxbjszolgwqvfiozofncbohduqgiswuiyddmwlwubetyaummenkdfptjczxemryuotrrymrfdxtrebpbjtpnuhsbnovhectpjhfhahbqrfbyxggobsweefcwxpqsspyssrmdhuelkkvyjxswjwofngpwfxvknkjviiavorwyfzlnktmfwxkvwkrwdcxjfzikdyswsuxegmhtnxjraqrdchaauazfhtklxsksbhwgjphgbasfnlwqwukprgvihntsyymdrfovaszjywuqygpvjtvlsvvqbvzsmgweiayhlubnbsitvfxawhfmfiatxvqrcwjshvovxknnxnyyfexqycrlyksderlqarqhkxyaqwlwoqcribumrqjtelhwdvaiysgjlvksrfvjlcaiwrirtkkxbwgicyhvakxgdjwnwmubkiazdjkfmotglclqndqjxethoutvjchjbkoasnnfbgrnycucfpeovruguzumgmgddqwjgdvaujhyqsqtoexmnfuluaqbxoofvotvfoiexbnprrxptchmlctzgqtkivsilwgwgvpidpvasurraqfkcmxhdapjrlrnkbklwkrvoaziznlpor",
data2 = "qhxepbshlrhoecdaodgpousbzfcqjxulatciapuftffahhlmxbufgjuxstfjvljybfxnenlacmjqoymvamphpxnolwijwcecgwbcjhgdybfffwoygikvoecdggplfohemfypxfsvdrseyhmvkoovxhdvoavsqqbrsqrkqhbtmgwaurgisloqjixfwfvwtszcxwktkwesaxsmhsvlitegrlzkvfqoiiwxbzskzoewbkxtphapavbyvhzvgrrfriddnsrftfowhdanvhjvurhljmpxvpddxmzfgwwpkjrfgqptrmumoemhfpojnxzwlrxkcafvbhlwrapubhveattfifsmiounhqusvhywnxhwrgamgnesxmzliyzisqrwvkiyderyotxhwspqrrkeczjysfujvovsfcfouykcqyjoobfdgnlswfzjmyucaxuaslzwfnetekymrwbvponiaojdqnbmboldvvitamntwnyaeppjaohwkrisrlrgwcjqqgxeqerjrbapfzurcwxhcwzugcgnirkkrxdthtbmdqgvqxilllrsbwjhwqszrjtzyetwubdrlyakzxcveufvhqugyawvkivwonvmrgnchkzdysngqdibhkyboyftxcvvjoggecjsajbuqkjjxfvynrjsnvtfvgpgveycxidhhfauvjovmnbqgoxsafknluyimkczykwdgvqwlvvgdmufxdypwnajkncoynqticfetcdafvtqszuwfmrdggifokwmkgzuxnhncmnsstffqpqbplypapctctfhqpihavligbrutxmmygiyaklqtakdidvnvrjfteazeqmbgklrgrorudayokxptswwkcircwuhcavhdparjfkjypkyxhbgwxbkvpvrtzjaetahmxevmkhdfyidhrdeejapfbafwmdqjqszwnwzgclitdhlnkaiyldwkwwzvhyorgbysyjbxsspnjdewjxbhpsvj",
hash = Object.create(null),
count = function (a) { hash[a] = (hash[a] || 0) + this; },
result;
data1.split('').forEach(count, 1);
data2.split('').forEach(count, -1);
result = Object.keys(hash).filter(function (k) { return hash[k]; });
console.log(result);
一种方法是对字符串进行排序,然后找出它们之间的第一个区别是:
var str = "1122334477422";
var str2 = "15122332247744";
var sorted = str.split('').sort();
var sorted2 = str2.split('').sort()
for(var i = 0; i< sorted2.length; i++){
if(i == sorted2.length || sorted[i] != sorted2[i]){
alert(sorted2[i]);
break;
}
}