Algorithm training(一)
2018-01-10 本文已影响8人
smart_yang
1. two sum
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
求目标值刚好是数组里面两个数字之和返回数组位置
解法:
/**
* two sum
*/
public static int[] twoSum(int[] numbers, int target) {
int[] result = new int[2];
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i = 0; i < numbers.length; i++) {
if (map.containsKey(target - numbers[i])) {//检查key值
result[1] = i + 1;
result[0] = map.get(target - numbers[i]);
return result;
}
map.put(numbers[i], i + 1);//存储数组的值作为key
}
return result;
}
2. Reverse Integer (数字 反转输出)
比如:
Input: 123
Output: 321
Input: -123
Output: -321
Input: 120
Output: 21
/**
* 反转数字 比如123 输出321
*/
public static int reverse(int x) {
int result = 0;
int sign = x >> 31;
try {
while (x != 0) {
result = result * 10 + x % 10;
x = x / 10;
}
if (sign != (result >> 31)) {
System.out.println("overflow..");
System.exit(1);
}
System.out.println("反转数字输出结果:" + result);
} catch (Exception e) {
e.printStackTrace();
}
return result;
}
避免不了溢出。。。
3.Remove Duplicates from Sorted Array (去掉分类数组里面相同值 返回array的长度)
Given nums = [1,1,2],里面相同1,返回长度2;
/**
* 去除相同 数字 返回数组长度
*
* @param nums{1,1,3,3,5,6,7}
* @return 数子值
*/
public static int removeDuplicates(int[] nums) {
if (nums.length == 0)
return 0;
int i = 0;
for (int j = 1; j < nums.length; j++) {
if (nums[j] != nums[i]) {
i++;
nums[i] = nums[j];
}
}
return i+1;
}
4.Remove Element
public int removeElement(int[] nums, int val) {
if(nums.length == 0) {
return 0;
}
int i = nums.length;
for (int j = 0; j < nums.length; j++) {
if (val == nums[j]) {
i--;
}
}
System.out.println("返回相同数字个数:" + i);
return i;
}
5.Implement strStr()
如果needle是haystack的一部分,返回索引值,如果不是,返回-1;
Input: haystack = "hello", needle = "ll"
Output: 2
Input: haystack = "aaaaa", needle = "bba"
Output: -1
实现:
public int strStr(String haystack, String needle) {
int l1 = haystack.length();
int l2 = needle.length();
if (l1 < l2) {
return -1;
} else if (l2 == 0) {
return 0;
}
int threshold = l1 - l2;
for (int i = 0; i <= threshold; ++i) {
if (haystack.substring(i, i + l2).equals(needle)) {
System.out.println("输出字符串位置:" + i);
return i;
}
}
return -1;
}