多试几次又有何妨,无论是恋爱或推理,今天或明天。
一. 单词距离问题
1 题目描述
给定一个字符串数组 wordDict 和两个已经存在于该数组中的字符串 word1 和 word2 。返回列表中这两个单词之间的最短距离。
示例 1:
输入: wordsDict = [“practice”, “makes”, “perfect”, “coding”, “makes”], word1 = “coding”, word2 = “practice”
输出: 3
示例 2:
输入: wordsDict = [“practice”, “makes”, “perfect”, “coding”, “makes”], word1 = “makes”, word2 = “coding”
输出: 1
2 解题思路
我们在一次遍历的过程中,获得两个目标单词的下标,然后计算其最小距离即可。
此题针对于word1和word2有两种情况:
- word1 != word2;
- word1和word2有可能相同;
3. 代码实现
3.1 针对word1 != word2的情况
public int shortestDistance(String[] wordsDict, String word1, String word2) {
if(wordsDict == null || wordsDict.length == 0) return 0;
int i1 = -1;
int i2 = -1;
int minDistance = wordsDict.length;
for(int i = 0; i < wordsDict.length; i++){
if(word1.equals(wordsDict[i])){
i1 = i;
if(i2 != -1){
minDistance = Math.min(minDistance,i1-i2);
}
}
if(word2.equals(wordsDict[i])){
i2 = i;
if(i1 != -1){
minDistance = Math.min(minDistance,i2-i1);
}
}
}
return minDistance;
}
3.2 针对word1可能等于word2的情况
我们只需要要过滤掉两个index指向同一个单词的情况(否则距离会被计算为0)
public int shortestDistance(String[] wordsDict, String word1, String word2) {
if(wordsDict == null || wordsDict.length == 0) return 0;
int i1 = -1;
int i2 = -1;
int minDistance = wordsDict.length;
for(int i = 0; i < wordsDict.length; i++){
if(word1.equals(wordsDict[i])){
i1 = i;
if(i2 != -1){
minDistance = Math.min(minDistance,i1-i2);
}
}
if(word2.equals(wordsDict[i])){
i2 = i;
// 注意这里的判断:i1 != i2
if(i1 != -1 && i1 != i2){
minDistance = Math.min(minDistance,i2-i1);
}
}
}
return minDistance;
}
PS: Java中字符串比较用 equals() 方法。
彩蛋
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