LeetCode 1. Two Sum--陣列中兩元素相加為該數值,輸出對應的兩個索引
阿新 • • 發佈:2019-01-05
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簡單來說,就是給一個數組及一個數值,求陣列中兩元素相加為該數值對應的兩個索引。且要求index1小於index2,而且答案結果穩定。, 1].
public class Solution { public int[] twoSum(int[] nums, int target) { int[] numsUnOrder = new int[nums.length]; for (int i = 0; i < nums.length; i++) {//原始無序陣列,不能直接使用int[] numsUnOrder = nums;這樣是引用傳遞,指向的還是nums[] numsUnOrder[i] = nums[i]; } int[] result = new int[2]; Arrays.sort(nums);//對陣列做了一次排序,預設升序排列 int begin = 0; int end = nums.length - 1; int temp; while (begin < end) { if (nums[begin] + nums[end] == target) { result[0] = nums[begin];//先存放數值 result[1] = nums[end]; break; } else if (nums[begin] + nums[end] < target) { begin++; } else if (nums[begin] + nums[end] > target) { end--; } }//while for (int i = 0; i < numsUnOrder.length; i++) {//從前往後比較,根據數值在原始無序陣列中查詢下標,並把下標存到result[]中 if (result[0] == numsUnOrder[i]) { result[0] = i; break; } } //可能會出現兩個相同的元素,所以需要一個從前往後查詢下標,一個從後往前查詢下標 for (int i = numsUnOrder.length - 1; i >= 0; i--) {//從後往前比較,根據數值在原始無序陣列中查詢下標,並把下標存到result[]中 if (result[1] == numsUnOrder[i]) { result[1] = i; break; } } if (result[0] > result[1]) {//保證result[0]比result[1]小 temp = result[0]; result[0] = result[1]; result[1] = temp; } return result; }//twoSum }
19 / 19 test cases passed.
Status: Accepted
Runtime: 7 ms
Your runtime beats 92.64 % of java submissions
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//AC 方法2 public int[] twoSum(int[] nums, int target) { int[] result = 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){ result[0] = i; result[1] = j; return result; } } } return result; }
19 / 19 test cases passed.
Status: Accepted
Runtime: 38 ms
Your runtime beats 29.73 % of java submissions.
時間複雜度是O(n^2),