一刷1509

Author: diguage

README.adoc

@@ -10589,12 +10589,12 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|Medium
 //|
 
-//|{counter:codes}
-//|{leetcode_base_url}/minimum-difference-between-largest-and-smallest-value-in-three-moves/[1509. Minimum Difference Between Largest and Smallest Value in Three Moves^]
-//|{source_base_url}/_1509_MinimumDifferenceBetweenLargestAndSmallestValueInThreeMoves.java[Java]
-//|{doc_base_url}/1509-minimum-difference-between-largest-and-smallest-value-in-three-moves.adoc[题解]
-//|Medium
-//|
+|{counter:codes}
+|{leetcode_base_url}/minimum-difference-between-largest-and-smallest-value-in-three-moves/[1509. Minimum Difference Between Largest and Smallest Value in Three Moves^]
+|{source_base_url}/_1509_MinimumDifferenceBetweenLargestAndSmallestValueInThreeMoves.java[Java]
+|{doc_base_url}/1509-minimum-difference-between-largest-and-smallest-value-in-three-moves.adoc[题解]
+|Medium
+|
 
 //|{counter:codes}
 //|{leetcode_base_url}/stone-game-iv/[1510. Stone Game IV^]

docs/1509-minimum-difference-between-largest-and-smallest-value-in-three-moves.adoc

@@ -1,69 +1,63 @@
 [#1509-minimum-difference-between-largest-and-smallest-value-in-three-moves]
-= 1509. Minimum Difference Between Largest and Smallest Value in Three Moves
+= 1509. 三次操作后最大值与最小值的最小差
 
-{leetcode}/problems/minimum-difference-between-largest-and-smallest-value-in-three-moves/[LeetCode - 1509. Minimum Difference Between Largest and Smallest Value in Three Moves ^]
+https://leetcode.cn/problems/minimum-difference-between-largest-and-smallest-value-in-three-moves/[LeetCode - 1509. 三次操作后最大值与最小值的最小差 ^]
 
-You are given an integer array `nums`.
+给你一个数组 `nums` 。
 
-In one move, you can choose one element of `nums` and change it to *any value*.
+每次操作你可以选择 `nums` 中的任意一个元素并将它改成 *任意值* 。
 
-Return _the minimum difference between the largest and smallest value of `nums` *after performing at most three moves*_.
+在 *执行最多三次移动后 *，返回 `nums` 中最大值与最小值的最小差值。
 
- 
-*Example 1:*
+*示例 1：*
 
-[subs="verbatim,quotes"]
-----
-*Input:* nums = [5,3,2,4]
-*Output:* 0
-*Explanation:* We can make at most 3 moves.
-In the first move, change 2 to 3. nums becomes [5,3,3,4].
-In the second move, change 4 to 3. nums becomes [5,3,3,3].
-In the third move, change 5 to 3. nums becomes [3,3,3,3].
-After performing 3 moves, the difference between the minimum and maximum is 3 - 3 = 0.
-
-----
-
-*Example 2:*
-
-[subs="verbatim,quotes"]
-----
-*Input:* nums = [1,5,0,10,14]
-*Output:* 1
-*Explanation:* We can make at most 3 moves.
-In the first move, change 5 to 0. nums becomes [1,0,0,10,14].
-In the second move, change 10 to 0. nums becomes [1,0,0,0,14].
-In the third move, change 14 to 1. nums becomes [1,0,0,0,1].
-After performing 3 moves, the difference between the minimum and maximum is 1 - 0 = 1.
-It can be shown that there is no way to make the difference 0 in 3 moves.
-----
-
-*Example 3:*
+....
+输入：nums = [5,3,2,4]
+输出：0
+解释：我们最多可以走 3 步。
+第一步，将 2 变为 3 。 nums 变成 [5,3,3,4] 。
+第二步，将 4 改为 3 。 nums 变成 [5,3,3,3] 。
+第三步，将 5 改为 3 。 nums 变成 [3,3,3,3] 。
+执行 3 次移动后，最小值和最大值之间的差值为 3 - 3 = 0 。
+....
 
-[subs="verbatim,quotes"]
-----
-*Input:* nums = [3,100,20]
-*Output:* 0
-*Explanation:* We can make at most 3 moves.
-In the first move, change 100 to 7. nums becomes [3,7,20].
-In the second move, change 20 to 7. nums becomes [3,7,7].
-In the third move, change 3 to 7. nums becomes [7,7,7].
-After performing 3 moves, the difference between the minimum and maximum is 7 - 7 = 0.
+*示例 2：*
 
-----
+....
+输入：nums = [1,5,0,10,14]
+输出：1
+解释：我们最多可以走 3 步。
+第一步，将 5 改为 0 。 nums变成 [1,0,0,10,14] 。
+第二步，将 10 改为 0 。 nums变成 [1,0,0,0,14] 。
+第三步，将 14 改为 1 。 nums变成 [1,0,0,0,1] 。
+执行 3 步后，最小值和最大值之间的差值为 1 - 0 = 1 。
+可以看出，没有办法可以在 3 步内使差值变为0。
+....
 
- 
-*Constraints:*
+*示例 3：*
 
+....
+输入：nums = [3,100,20]
+输出：0
+解释：我们最多可以走 3 步。
+第一步，将 100 改为 7 。 nums 变成 [3,7,20] 。
+第二步，将 20 改为 7 。 nums 变成 [3,7,7] 。
+第三步，将 3 改为 7 。 nums 变成 [7,7,7] 。
+执行 3 步后，最小值和最大值之间的差值是 7 - 7 = 0。
+....
 
-* `1 <= nums.length <= 10^5^`
-* `-10^9^ <= nums[i] <= 10^9^`
 
+*提示：*
 
+* `1 \<= nums.length \<= 10^5^`
+* `-10^9^ \<= nums[i] \<= 10^9^`
 
 
 == 思路分析
 
+先对数组进行排序，要求最大值与最小值的最小差值，那么就是把减少最大值和最小值直接的差值，要么把最大值变小，要么把最小值变大，或者两者都要。所以，就要看那种方式能对差值改变最大。
+
+看官方题解，可以直接求 stem:[a(n - 4 + i) - a(i)] 的最小值。
 
 [[src-1509]]
 [tabs]
@@ -90,4 +84,4 @@ include::{sourcedir}/_1509_MinimumDifferenceBetweenLargestAndSmallestValueInThre
 
 == 参考资料
 
-
+. https://leetcode.cn/problems/minimum-difference-between-largest-and-smallest-value-in-three-moves/solutions/336428/san-ci-cao-zuo-hou-zui-da-zhi-yu-zui-xiao-zhi-de-2/[1509. 三次操作后最大值与最小值的最小差 - 官方题解^]

docs/index.adoc

@@ -3101,7 +3101,7 @@ include::1480-running-sum-of-1d-array.adoc[leveloffset=+1]
 
 // include::1508-range-sum-of-sorted-subarray-sums.adoc[leveloffset=+1]
 
-// include::1509-minimum-difference-between-largest-and-smallest-value-in-three-moves.adoc[leveloffset=+1]
+include::1509-minimum-difference-between-largest-and-smallest-value-in-three-moves.adoc[leveloffset=+1]
 
 // include::1510-stone-game-iv.adoc[leveloffset=+1]
 

logbook/202503.adoc

@@ -763,6 +763,11 @@ endif::[]
 |{doc_base_url}/3111-minimum-rectangles-to-cover-points.adoc[题解]
 |✅ 贪心。从左边界开始，尽可能多占用横轴。宽度耗尽，则从下一个点开始。
 
+|{counter:codes}
+|{leetcode_base_url}/minimum-difference-between-largest-and-smallest-value-in-three-moves/[1509. 三次操作后最大值与最小值的最小差^]
+|{doc_base_url}/1509-minimum-difference-between-largest-and-smallest-value-in-three-moves.adoc[题解]
+|✅ 贪心。转变一下题目就是要求如何缩短最大值和最小值直接的差值。看官方题解，可以直接求 stem:[a(n - 4 + i) - a(i)] 的最小值。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

src/main/java/com/diguage/algo/leetcode/_1509_MinimumDifferenceBetweenLargestAndSmallestValueInThreeMoves.java

@@ -0,0 +1,34 @@
+package com.diguage.algo.leetcode;
+
+import java.util.Arrays;
+
+public class _1509_MinimumDifferenceBetweenLargestAndSmallestValueInThreeMoves {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-05-21 22:54:47
+   */
+  public int minDifference(int[] nums) {
+    int length = nums.length;
+    if (length <= 4) {
+      return 0;
+    }
+    Arrays.sort(nums);
+    int diff = nums[length - 1] - nums[0];
+    int max = 0;
+    // max = Math.max(max, (nums[3] - nums[0]) + (nums[length - 1] - nums[length - 1]));
+    // max = Math.max(max, (nums[2] - nums[0]) + (nums[length - 1] - nums[length - 2]));
+    // max = Math.max(max, (nums[1] - nums[0]) + (nums[length - 1] - nums[length - 3]));
+    // max = Math.max(max, (nums[0] - nums[0]) + (nums[length - 1] - nums[length - 4]));
+    for (int i = 3; i >= 0; i--) {
+      max = Math.max(max, (nums[i] - nums[0]) + (nums[length - 1] - nums[length - (4 - i)]));
+    }
+    return diff - max;
+  }
+  // end::answer[]
+  public static void main(String[] args) {
+    new _1509_MinimumDifferenceBetweenLargestAndSmallestValueInThreeMoves()
+      .minDifference(new int[]{1, 5, 0, 10, 14});
+  }
+
+}