From 205a1532da4a3e0973ce6149568d31c9e5d6d59a Mon Sep 17 00:00:00 2001
From: lichaoxi <lichaoxi7@qq.com>
Date: Wed, 1 Jan 2020 10:30:04 +0800
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原本实现，在计算 matrix[i][j] <= mid 的 count 时，耗时较高
---
 notes/Leetcode 题解 - 数组与矩阵.md | 28 +++++++++++++++++-----------
 1 file changed, 17 insertions(+), 11 deletions(-)

diff --git a/notes/Leetcode 题解 - 数组与矩阵.md b/notes/Leetcode 题解 - 数组与矩阵.md
index 288147f1..3560f3ee 100644
--- a/notes/Leetcode 题解 - 数组与矩阵.md	
+++ b/notes/Leetcode 题解 - 数组与矩阵.md	
@@ -146,20 +146,26 @@ return 13.
 
 ```java
 public int kthSmallest(int[][] matrix, int k) {
-    int m = matrix.length, n = matrix[0].length;
-    int lo = matrix[0][0], hi = matrix[m - 1][n - 1];
-    while (lo <= hi) {
-        int mid = lo + (hi - lo) / 2;
-        int cnt = 0;
-        for (int i = 0; i < m; i++) {
-            for (int j = 0; j < n && matrix[i][j] <= mid; j++) {
-                cnt++;
+    int n = matrix.length;
+    int low = matrix[0][0], high = matrix[n-1][n-1];
+    while (low < high) {
+        int mid = (low + high) >> 1;
+        int count = 0;
+        for (int i = 0, j = n - 1; i < n && j >= 0;) {
+            if (matrix[i][j] <= mid) {
+                count += j + 1;
+                i++;
+            } else {
+                j--;
             }
         }
-        if (cnt < k) lo = mid + 1;
-        else hi = mid - 1;
+        if (count < k) {
+            low = mid + 1;
+        } else {
+            high = mid;
+        }
     }
-    return lo;
+    return high;
 }
 ```