LeetCode之Find the Difference
【摘要】 1、题目
Given two strings s and t which consist of only lowercase letters.
String t is generated by random shuffling string s and then add one mor...
1、题目
Given two strings s and t which consist of only lowercase letters.
String t is generated by random shuffling string s and then add one more letter at a random position.
Find the letter that was added in t.
Example:
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Input:
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s = "abcd"
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t = "abcde"
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Output:
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e
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Explanation:
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'e' is the letter that was added.
Input:
s = "a"
t = "aa"
Output:
a
2、代码实现
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public class Solution {
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public static char findTheDifference(String s, String t) {
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if (s == null || t.length() == 0)
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return t.charAt(0);
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if (t == null || t.length() == 0)
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return s.charAt(0);
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if (s == null && t == null)
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return 0;
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int[] a = new int[30];
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char[] tChars = t.toCharArray();
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char[] sChars = s.toCharArray();
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int sLength = s.length();
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int tLength = t.length();
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if (sLength > tLength) {
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for (int i = 0; i < sChars.length; i++) {
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if (a[sChars[i] - 97] != 0)
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a[sChars[i] - 97] = ++(a[sChars[i] - 97]);
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else
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a[sChars[i] - 97] = 2;
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}
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for (int i = 0; i < tChars.length; i++) {
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a[tChars[i] - 97] = --(a[tChars[i] - 97]);
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}
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} else {
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for (int i = 0; i < tChars.length; i++) {
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if (a[tChars[i] - 97] != 0)
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a[tChars[i] - 97] = ++(a[tChars[i] - 97]);
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else
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a[tChars[i] - 97] = 2;
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}
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for (int i = 0; i < sChars.length; i++) {
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a[sChars[i] - 97] = --(a[sChars[i] - 97]);
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}
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}
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for (int i = 0; i < 30; i ++) {
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if (a[i] >= 2) {
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return (char) (i + 97);
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}
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}
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return 0;
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}
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}
3、总结
看到2个字符串对比，我么可以先转化为字符数组，下表从A - 65 活着 a - 95 开始，也就是从下表0开始，然后要注意2个字符串里面可能包含同样的元素有几个的情况，相同就往上加，另外一个就减，但是他们最多相差1个字符，所以，我们可可以肯定，比我们一开始设置的大，也就是3，然后如果没有重复的数据，那么一样的就为1，肯定有一个为2.
文章来源: chenyu.blog.csdn.net，作者：chen.yu，版权归原作者所有，如需转载，请联系作者。
原文链接：chenyu.blog.csdn.net/article/details/67845937
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