總結(反省):
1、不知道如何下手,沒看懂要求
實際上只是寫出求解問題的方法,參數之類的在方法參數列表中已傳。
2、數組的長度arr.length,已經模糊
3、for循環從0開始,竟然到arr.length結束
for(int i = 0; i <= arr.length;i++){
...
}
4、方法返回值的忽略
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以下是自己與答案之間的對比(感覺自己很差勁):
自己的:
public int[] twoSum(int[] nums, int target) {
int[] arr = new int[2];
for(int i = 0;i < nums.length;i++){
for(int j = i + 1;j < nums.length ; j++){
if(nums[i]+nums[j]==target){
arr[0] = i;
arr[1] = j;
}
}
}
return arr;
}
答案:
public int[] twoSum(int[] nums, int target) { for (int i = 0; i < nums.length; i++) { for (int j = i + 1; j < nums.length; j++) { if (nums[j] == target - nums[i]) { return new int[] { i, j }; } } } throw new IllegalArgumentException("No two sum solution"); }
不僅如此,上面的答案只是一種蠻力(Brute Force)求解的過程,根本沒有編程之美( ∩_∩ ,自我狼狽中~~)
這裏給個好的(One-pass Hash Table):
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement)) {
return new int[] { map.get(complement), i };
}
map.put(nums[i], i);
}
throw new IllegalArgumentException("No two sum solution");
}
下面給出鏈接 可以自己去體驗
https://leetcode.com/articles/two-sum/
-----------------------------------------------------------------------------------------------------------------------------------原題:
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
翻譯:
給定一組整數,將兩個數的返回指數相加,使它們相加成一個特定的目標。
您可以假設每個輸入都有一個解決方案,您可能不會使用相同的元素兩次。