diff --git a/Algorithms/0475. Heaters/475. Heaters.go b/Algorithms/0475. Heaters/475. Heaters.go
new file mode 100644
index 00000000..381a89a9
--- /dev/null
+++ b/Algorithms/0475. Heaters/475. Heaters.go	
@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"math"
+	"sort"
+)
+
+func findRadius(houses []int, heaters []int) int {
+	minRad := 0
+	sort.Ints(heaters)
+	for _, house := range houses {
+		// 遍历房子的坐标，维护 heaters 的最小半径
+		heater := findClosestHeater(house, heaters)
+		rad := heater - house
+		if rad < 0 {
+			rad = -rad
+		}
+		if rad > minRad {
+			minRad = rad
+		}
+	}
+	return minRad
+
+}
+
+// 二分搜索
+func findClosestHeater(pos int, heaters []int) int {
+	low, high := 0, len(heaters)-1
+	if pos < heaters[low] {
+		return heaters[low]
+	}
+	if pos > heaters[high] {
+		return heaters[high]
+	}
+	for low <= high {
+		mid := low + (high-low)>>1
+		if pos == heaters[mid] {
+			return heaters[mid]
+		} else if pos < heaters[mid] {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	// 判断距离两边的 heaters 哪个更近
+	if pos-heaters[high] < heaters[low]-pos {
+		return heaters[high]
+	}
+	return heaters[low]
+}
+
+// 解法二 暴力搜索
+func findRadius1(houses []int, heaters []int) int {
+	res := 0
+	for i := 0; i < len(houses); i++ {
+		dis := math.MaxInt64
+		for j := 0; j < len(heaters); j++ {
+			dis = min(dis, abs(houses[i]-heaters[j]))
+		}
+		res = max(res, dis)
+	}
+	return res
+}
diff --git a/Algorithms/0475. Heaters/475. Heaters_test.go b/Algorithms/0475. Heaters/475. Heaters_test.go
new file mode 100644
index 00000000..f39489ee
--- /dev/null
+++ b/Algorithms/0475. Heaters/475. Heaters_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question475 struct {
+	para475
+	ans475
+}
+
+// para 是参数
+// one 代表第一个参数
+type para475 struct {
+	houses  []int
+	heaters []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans475 struct {
+	one int
+}
+
+func Test_Problem475(t *testing.T) {
+
+	qs := []question475{
+
+		question475{
+			para475{[]int{1, 2, 3}, []int{2}},
+			ans475{1},
+		},
+
+		question475{
+			para475{[]int{1, 2, 3, 4}, []int{1, 4}},
+			ans475{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 475------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans475, q.para475
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findRadius(p.houses, p.heaters))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0475. Heaters/README.md b/Algorithms/0475. Heaters/README.md
new file mode 100755
index 00000000..a0375170
--- /dev/null
+++ b/Algorithms/0475. Heaters/README.md	
@@ -0,0 +1,51 @@
+# [475. Heaters](https://leetcode.com/problems/heaters/)
+
+## 题目:
+
+Winter is coming! Your first job during the contest is to design a standard heater with fixed warm radius to warm all the houses.
+
+Now, you are given positions of houses and heaters on a horizontal line, find out minimum radius of heaters so that all houses could be covered by those heaters.
+
+So, your input will be the positions of houses and heaters seperately, and your expected output will be the minimum radius standard of heaters.
+
+**Note:**
+
+1. Numbers of houses and heaters you are given are non-negative and will not exceed 25000.
+2. Positions of houses and heaters you are given are non-negative and will not exceed 10^9.
+3. As long as a house is in the heaters' warm radius range, it can be warmed.
+4. All the heaters follow your radius standard and the warm radius will the same.
+
+**Example 1:**
+
+    Input: [1,2,3],[2]
+    Output: 1
+    Explanation: The only heater was placed in the position 2, and if we use the radius 1 standard, then all the houses can be warmed.
+
+**Example 2:**
+
+    Input: [1,2,3,4],[1,4]
+    Output: 1
+    Explanation: The two heater was placed in the position 1 and 4. We need to use radius 1 standard, then all the houses can be warmed.
+
+
+
+## 题目大意
+
+
+冬季已经来临。 你的任务是设计一个有固定加热半径的供暖器向所有房屋供暖。现在，给出位于一条水平线上的房屋和供暖器的位置，找到可以覆盖所有房屋的最小加热半径。所以，你的输入将会是房屋和供暖器的位置。你将输出供暖器的最小加热半径。
+
+说明:
+
+- 给出的房屋和供暖器的数目是非负数且不会超过 25000。
+- 给出的房屋和供暖器的位置均是非负数且不会超过10^9。
+- 只要房屋位于供暖器的半径内(包括在边缘上)，它就可以得到供暖。
+- 所有供暖器都遵循你的半径标准，加热的半径也一样。
+
+
+
+## 解题思路
+
+
+- 给出一个房子坐标的数组，和一些供暖器坐标的数组，分别表示房子的坐标和供暖器的坐标。要求找到供暖器最小的半径能使得所有的房子都能享受到暖气。
+- 这一题可以用暴力的解法，暴力解法依次遍历每个房子的坐标，再遍历每个供暖器，找到距离房子最近的供暖器坐标。在所有这些距离的长度里面找到最大值，这个距离的最大值就是供暖器半径的最小值。时间复杂度 O(n^2)。
+- 这一题最优解是二分搜索。在查找距离房子最近的供暖器的时候，先将供暖器排序，然后用二分搜索的方法查找。其他的做法和暴力解法一致。时间复杂度 O(n log n)。