Contains Duplicate II
思路一: for 循环遍历,结果超时
public boolean containsNearbyDuplicate(int[] nums, int k) {
int left = -1;
for (int i = 0; i < nums.length; i++) {
left = i;
for (int j = i + 1; j < nums.length; j++) {
if (nums[left] == nums[j]) {
if (Math.abs(left - j) <= k) return true;
else left = j;
}
}
}
return false;
}
思路二: 用 map 记录数组下标信息,遇到相等的数字取出下标对比
public boolean containsNearbyDuplicate(int[] nums, int k) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
if (map.containsKey(nums[i])) {
if (Math.abs(map.get(nums[i]) - i) <= k) return true;
else map.put(nums[i], i);
} else {
map.put(nums[i], i);
}
}
return false;
}
标签:map,nums,int,219,++,length,easy,leetcode,left
From: https://www.cnblogs.com/iyiluo/p/16804876.html