LeetCode第28题
Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
Example 1:
Input: haystack = "hello", needle = "ll"
Output: 2
Example 2:
Input: haystack = "aaaaa", needle = "bba"
Output: -1
翻译：
返回大字符串中第一次找到给定小字符串的下标
思路：
String有自带API：indexOf
代码：
class Solution {
public int strStr(String haystack, String needle) {
return haystack.indexOf(needle);
}
}
我们来看下SDK源码
public int indexOf(String str) {
return indexOf(str, 0);
}
public int indexOf(String str, int fromIndex) {
return indexOf(this, str, fromIndex);
}
static int indexOf(String source,
String target,
int fromIndex) {
if (fromIndex >= source.count) {
return (target.count == 0 ? source.count : -1);
}
if (fromIndex < 0) {
fromIndex = 0;
}
if (target.count == 0) {
return fromIndex;
}
char first = target.charAt(0);
int max = (source.count - target.count);
for (int i = fromIndex; i <= max; i++) {
/* Look for first character. */
if (source.charAt(i)!= first) {
while (++i <= max && source.charAt(i) != first);
}
/* Found first character, now look at the rest of v2 */
if (i <= max) {
int j = i + 1;
int end = j + target.count - 1;
for (int k = 1; j < end && source.charAt(j)
== target.charAt(k); j++, k++);
if (j == end) {
/* Found whole string. */
return i;
}
}
}
return -1;
}
思路也很简单和暴力，就是遍历对比字符
LeetCode第35题
Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.
Example 1:
Input: [1,3,5,6], 5
Output: 2
Example 2:
Input: [1,3,5,6], 2
Output: 1
Example 3:
Input: [1,3,5,6], 7
Output: 4
Example 4:
Input: [1,3,5,6], 0
Output: 0
翻译：
给定一个有序数组和一个给定值，在数组中找到这个值所在的下标；如果没有这个值，那么返回他应该插入的下标位置
思路：
和选择排序或插入排序思路相似，就是将定值和已经排好序的数组遍历比较大小
代码：
class Solution {
public int searchInsert(int[] nums, int target) {
int j= 0;
for(int i = 0;i<nums.length;i++){
if(nums[i] < target){
j++;
}
}
return j;
}
}
如果数组元素比给定值要小，那么J就+1；如果数组元素比给定值要大，J不操作；那么下标比J小的元素值就比给定值要小，即J就是我们所需要的下标
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