coins problem DP & BackTracking solution

python/41_dynamic_programming/coins_problem.py

@@ -0,0 +1,45 @@
+#!/usr/bin/python
+# -*- coding: UTF-8 -*-
+
+from typing import List
+
+
+def coins_dp(values: List[int], target: int) -> int:
+    # memo[i]表示target为i的时候，所需的最少硬币数
+    memo = [0] * (target+1)
+    # 0元的时候为0个
+    memo[0] = 0
+
+    for i in range(1, target+1):
+        min_num = 999999
+        # 对于values中的所有n
+        # memo[i]为 min(memo[i-n1], memo[i-n2], ...) + 1
+        for n in values:
+            if i >= n:
+                min_num = min(min_num, 1 + memo[i-n])
+            else:   # values中的数值要从小到大排序
+                break
+        memo[i] = min_num
+
+    # print(memo)
+    return memo[-1]
+
+
+min_num = 999999
+def coins_backtracking(values: List[int], target: int, cur_value: int, coins_count: int):
+    if cur_value == target:
+        global min_num
+        min_num = min(coins_count, min_num)
+    else:
+        for n in values:
+            if cur_value + n <= target:
+                coins_backtracking(values, target, cur_value+n, coins_count+1)
+
+
+if __name__ == '__main__':
+    values = [1, 3, 5]
+    target = 23
+    print(coins_dp(values, target))
+    coins_backtracking(values, target, 0, 0)
+    print(min_num)
+