1. Two Sum

暴力解法O(n2)

略

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java中map的get时间为O(1) C++中为O（logn）

哈希解法第一版 O(n）

public class Solution {
public int[] twoSum(int[] nums, int target) {
int[] results = new int[2];
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i = 0; i < nums.length; map.put(nums[i], i++)) {
if(map.containsKey(target - nums[i])) {
results[0] = map.get(target - nums[i]);
results[1] = map.get(nums[i]);
return results;
}
}
return results;
}
}

循环中results[1] = map.get(nums[i])为NULL，注意大小写

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哈希解法

public class Solution {
public int[] twoSum(int[] nums, int target) {
int[] results = new int[2];
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i = 0; i < nums.length; map.put(nums[i], i++)) {
if(map.containsKey(target - nums[i])) {
results[0] = map.get(target - nums[i]);
results[1] = i;
return results;
}
}
return results;
}
}

转载于:https://www.cnblogs.com/bloomingFlower/p/6789692.html