242-e-Valid Anagram
2015-08-02 16:42
330 查看
判断字母替换是否相同。一看就是hash了,用一个26位数组就行了,过s时加1过t时减1,最后判断数组是否全0就行。最近感觉leet新出的easy题没有以前的难了。
如下,做了大小写判断:
bool isAnagram(char* s, char* t) {
bool result = true;
int n = 26;
char hmap
;
memset(hmap, 0, sizeof(char) * n);
int index_s = 0, index_t = 0;
while (*s != '\0') { //s
index_s = *s - 'A';
if (*s >= 'a' && *s <= 'z')
index_s -= 'a' - 'A';
hmap[index_s]++;
s++;
}
while (*t != '\0') {
index_t = *t - 'A';
if (*t >= 'a' && *t <= 'z')
index_t -= 'a' - 'A';
hmap[index_t]--;
t++;
}
for (int i = 0; i < n; i++)
if (hmap[i] != 0) {
result = false;
break;
}
return result;
}
如下,做了大小写判断:
bool isAnagram(char* s, char* t) {
bool result = true;
int n = 26;
char hmap
;
memset(hmap, 0, sizeof(char) * n);
int index_s = 0, index_t = 0;
while (*s != '\0') { //s
index_s = *s - 'A';
if (*s >= 'a' && *s <= 'z')
index_s -= 'a' - 'A';
hmap[index_s]++;
s++;
}
while (*t != '\0') {
index_t = *t - 'A';
if (*t >= 'a' && *t <= 'z')
index_t -= 'a' - 'A';
hmap[index_t]--;
t++;
}
for (int i = 0; i < n; i++)
if (hmap[i] != 0) {
result = false;
break;
}
return result;
}
相关文章推荐
- leetcode 179 Largest Number
- leetcode 24 Swap Nodes in Pairs
- leetcode 2 Add Two Numbers 方法1
- leetcode 2 Add Two Numbers 方法2
- [LeetCode]47 Permutations II
- [LeetCode]65 Valid Number
- [LeetCode]123 Best Time to Buy and Sell Stock III
- [LeetCode] String Reorder Distance Apart
- [LeetCode] Sliding Window Maximum
- [LeetCode] Find the k-th Smallest Element in the Union of Two Sorted Arrays
- [LeetCode] Determine If Two Rectangles Overlap
- [LeetCode] A Distance Maximizing Problem
- leetcode_linearList
- leetcode_linearList02
- LeetCode[Day 1] Two Sum 题解
- LeetCode[Day 2] Median of Two Sorted Arrays 题解
- LeetCode[Day 3] Longest Substring Without... 题解
- LeetCode [Day 4] Add Two Numbers 题解
- LeetCode [Day 5] Longest Palindromic Substring 题解
- LeetCode [Day 6] ZigZag Conversion 题解