标签：find pre problems return == problem 复杂度 hash nta
思路很简单这里就不说了，下面贴了不同的几个Java实现的时间与其他算法实现的时间的比较
这个是LeetCode的第一道题，也是我刷的第一道，刚一开始的Java实现
public int[] twoSum(int[] nums, int target) {
boolean find = false;
int targeti = -1, targetj = -1;
for (int i = 0; i < nums.length; i++) {
for (int j = 0; j < nums.length; j++) {
if (i == j) {
continue;
}
if (nums[i] + nums[j] == target) {
targeti = i;
targetj = j;
find = true;
break;
}
}
if (find) {
break;
}
}
return new int[]{targeti, targetj};
}
稍微改进一下
public int[] twoSum(int[] nums, int target) {
boolean find = false;
int targeti = -1, targetj = -1;
for (int i = 0; i < nums.length - 1; i++) {
for (int j = i + 1; j < nums.length; j++) {
if (nums[i] + nums[j] == target) {
targeti = i;
targetj = j;
find = true;
break;
}
}
if (find) {
break;
}
}
return new int[]{targeti, targetj};
}
下面这个版本是在讨论中看到的O(1)时间复杂度
public int[] twoSum3(int[] nums, int target) {
int[] result = new int[2];
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i = 0; i < nums.length; i++) {
if (map.containsKey(target - nums[i])) {
result[1] = i;
result[0] = map.get(target - nums[i]);
return result;
}
map.put(nums[i], i);
}
return result;
}
标签：find pre problems return == problem 复杂度 hash nta
原文地址：https://www.cnblogs.com/okokabcd/p/9191871.html
