LeetCode 第一题,两数之和
给定 nums = [2, 7, 11, 15], target = 9
因为 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]
解题:
暴力解决法,两个 for 循环,时间复杂度
O(n2)
map 存储差值,遍历时检查差值是否存在,存在则返回,不存在
set
,时间复杂度O(n)
- JavaScript
var twoSum = function(nums, target) {
const map = new Map();
for (let i = 0; i < nums.length; i++) {
let v = target - nums[i];
if (map.has(v)) {
return [map.get(v), i]
} else {
map.set(nums[i], i);
}
}
};
- Go
func twoSum(nums []int, target int) []int {
var result []int
sumMap := make(map[int]int)
for i := 0; i < len(nums); i++ {
v := target - nums[i]
index, ok := sumMap[v]
if ok {
result = []int{index, i}
break
} else {
sumMap[nums[i]] = i
}
}
return result
}