feat(233): ✨数位 DP 解决 233

Author: descire

Rank/182/solution4.js renamed to Rank/182/5371/solution1.js

@@ -6,15 +6,24 @@ const findGoodStrings = (n, s1, s2, evil) => {
     return 0;
   }
 
+  const record = new Set();
+  for (let i = 0; i < evil.length; i++) {
+    record.add(evil[i]);
+  }
+
   let c1 = 0
   for (let i = s1[0].charCodeAt(0) + 1; i < s2[0].charCodeAt(0); i++) {
     if (evil.indexOf(String.fromCharCode(i)) === -1) {
       c1++;
     }
   }
+
   ans += getCount(c1, evil.length, n - 1);
 
   for (let i = 1; i < s1.length; i++) {
+    if (record.has(s1[i - 1])) {
+      break;
+    }
     const item = s1[i];
     let count = 0;
     let start = item.charCodeAt(0);
@@ -26,7 +35,13 @@ const findGoodStrings = (n, s1, s2, evil) => {
       start++;
     }
     ans += getCount(count, evil.length, n - i - 1);
+      
+  }
 
+  for (let i = 0; i < s2.length; i++) {
+    if (record.has(s2[i - 1])) {
+      break;
+    }
     if(s2[i - 1] !== s1[i - 1]) {
       const item2 = s2[i];
       let count2 = 0;
@@ -40,8 +55,8 @@ const findGoodStrings = (n, s1, s2, evil) => {
 
       ans += getCount(count2, evil.length, n - i - 1);
     }
-      
   }
+
   return ans;
 }
 

dynamic-programming/233/solution1.js

@@ -0,0 +1,50 @@
+/**
+ * https://leetcode-cn.com/problems/number-of-digit-one/
+ * 
+ * 233. 数字 1 的个数
+ * 
+ * Hard
+ */
+const countDigitOne = n => {
+  if (n <= 0) {
+    return 0;
+  }
+
+  if (n === 1) {
+    return 1;
+  }
+
+  const list = String(n).split('');
+  let ans = 0;
+  for (let i = 0, max = list.length; i < max; i++) {
+    const currentItem = list[i];
+
+    if (currentItem > 1) {
+      let leftCount = Number.parseInt(list.slice(0, i).join('')) + 1;
+      if (Number.isNaN(leftCount)) {
+        leftCount = 1;
+      }
+      ans += leftCount * Math.pow(10, max - i - 1);
+    } else if (currentItem === '1') {
+      let leftCount = Number.parseInt(list.slice(0, i).join(''));
+      if (Number.isNaN(leftCount)) {
+        leftCount = 0;
+      }
+      ans += leftCount * Math.pow(10, max - i - 1);
+      let rightCount = Number.parseInt(list.slice(i + 1).join(''));
+      if (Number.isNaN(rightCount)) {
+        rightCount = 0;
+      }
+      ans += rightCount + 1;
+    } else {
+      let leftCount = Number.parseInt(list.slice(0, i).join(''));
+      if (Number.isNaN(leftCount)) {
+        leftCount = 1;
+      }
+      ans += leftCount * Math.pow(10, max - i - 1);
+    }
+  
+  }
+
+  return ans;
+}